Ch. 7: Z-transform Reading
|
|
- Elfrieda Austin
- 6 years ago
- Views:
Transcription
1 c J. Fessler, June 9, 3, 6:3 (student version) 7. Ch. 7: Z-transform Definition Properties linearity / superposition time shift convolution: y[n] =h[n] x[n] Y (z) =H(z) X(z) Inverse z-transform by coefficient matching System function H(z) poles, zeros, pole-zero plots conjugate pairs relationship to H(ˆω) Interconnection of systems Cascade / series connection Parallel connection Feedback connection Filter design Reading Text Ch. 7 (Section 7.9 optional)
2 c J. Fessler, June 9, 3, 6:3 (student version) 7. z-transforms Introduction So far we have discussed the time domain (n), and the frequency domain (ˆω). We now turn to the z domain (z). Why? Other types of input signals: step functions, geometric series, finite-duration sinusoids. Transient analysis. Right now we only have a time-domain approach. Filter design: approaching a systematic method. Analysis and design of IIR filters. Definition The (one-sided) z transform of a signal x[n] is defined by X(z) = x[k] z k = x[] + x[] z + x[] z +, k= where z can be any complex number. This is a function of z. We write X(z) =Z{x[n]}, where Z{ }denotes the z-transform operation. We call x[n] and X(z) z-transform pairs, denoted x[n] X(z). Example. For the signal the z-transform is Left side: function of time (n) Right side: function of z 3, n = x[n] =3δ[n]+7δ[n 6] = 7, n =6, otherwise, X(z) = k (3δ[k]+7δ[k 6])z k =3+z 6. This is a polynomial in z ; specifically: X(z) =3+7(z ) 6. n-domain z-domain x[n] = k x[k] δ[n k] X(z) = k x[k] z k For causal signals, the z-transform is one-to-one, so using coefficient matching we can determine the signal x[n] from its z- transform. We call this taking the inverse z-transform. Example. Given Y (z) =3z +7z 5, the corresponding signal is y[n] =3δ[n ] + 7δ[n 5]. For infinite duration signals there are some technicalities that we leave to EECS 36 and 45.
3 c J. Fessler, June 9, 3, 6:3 (student version) 7.3 Properties of the z-transform These are all easy to show from the definition of the z-transform. Linearity αx[n] +βy[n] αx(z) +βy (z) Time-shift Convolution (later) x[n n ] X(z) z n y[n] =x[n] h[n] Y (z) =H(z) X(z) The system function of an LTI system Summary of ways to characterize filters. Time domain: I/O relation / diffeq y[n] = Filter coefficients {b k } Impulse response h[n] Block diagram Frequency domain: Frequency response H(ˆω) System function H(z) Pole-zero plot (z domain poles, zeros, gain) The z-transform is particularly useful when we consider LTI systems. The z-transform of the impulse response h[n] of an LTI system is denoted H(z) and is called the system function. We will focus on causal systems, so the system function, also called the transfer function,isgivenby H(z) = h[k] z k = h[] + h[] z + h[] z +. k= Example. The two-point moving average filter has impulse response h[n] = δ[n]+ δ[n ]. Its system function is H(z) = + z. H(z) Magnitude Response H(ω) π π/ ω π/ π
4 c J. Fessler, June 9, 3, 6:3 (student version) 7.4 Example. A filter has the system function H(z) =z 3. What does it do? Taking the inverse z-transform by coefficient matching we see so this system delays the signal by 3 samples. h[n] =δ[n 3], This is why a unit delay is often denoted by z in block diagrams. Example. Thez-transform is particularly useful for IIR filters. Suppose h[n] =a n u[n], for some number a with a <. Then H(z) = h[k] z k = a k z k = (az ) k = k= k= k= az. Does this hold for all values of z? No! We need to have az < i.e., z > a. The subset of the complex plane {z C : z > a } is called the region of convergence of the z-transform. More details in 36/45. Example. A filter has impulse response h[n] =δ[n]+(/) n u[n]. The system function is H(z) =+ z = z z. Notice that this is a ratio of polynomials in z. We call such system functions rational system functions, and they are the only kind we will need for designing FIR and IIR filters. The z domain versus the ˆω domain The z domain is a generalization of the frequency domain for causal signals such as suddenly-applied sinusoids. Notice the similarity: Frequency response of a causal LTI system: H(ˆω) = System function for a causal LTI system: H(z) = k= h[k]e j ˆωk. k= h[k] z k. Comparing, we see that if we replace z with e j ˆω (which is a complex value after all for any given ˆω) in the system function, then we obtain the frequency response: H(ˆω) =H(z) = H ( e j ˆω). z=e j ˆω This is why we use the notation H(ˆω) and H(z), because they are very closely related, by making the substitution z =e j ˆω. For such values of z, wehave z = e j ˆω =and z = e j ˆω =ˆω, so these values of z lie on the unit circle in the complex plane. So H(z) is more general because one can also consider values of z that are not on the unit circle, and this flexibility is helpful for filter design. Furthermore, z is easier to write than e j ˆω! Example. What is the magnitude response of the filter with system function H(z) =z 3?? H(ˆω) =H ( e j ˆω) =e j 3ˆω, so H(ˆω) = e j 3ˆω =. So the 3-sample delay system is an all pass filter. The magnitudes of sinusoidal components are unaffected. Where is DC? At ˆω =,soz =. Where are high frequencies? Near ˆω = ±π,soz =. The frequency interval π ˆω π corresponds to the entire unit circle. For eternal periodic signals, the z-transform is undefined, so the frequency domain remains our primary tool for those signals.
5 c J. Fessler, June 9, 3, 6:3 (student version) 7.5 Summary for FIR filters Time domain: Frequency response (ˆω): System function (z): Avoiding convolution Time domain (convolution): z domain (multiplication!): M M y[n] = b k x[n k], h[n] = b k δ[n k] k= H(ˆω) = H(z) = M k= j ˆωk b k e M b k z k. k= k= x[n] h[n] y[n] =x[n] h[n] X(z) H(z) Y (z) =X(z) H(z) Example. (Aperiodic input signal, so H(ˆω) is not directly useful!) x[n] =3δ[n]+δ[n ] H(ˆω) =e j ˆω e j ˆω y[n] =? To avoid convolution, use z domain. X(z) =3+z,H(z) =z + z, so Taking the inverse z-transform yields Y (z) =X(z) H(z) =(3+z )(z + z )=3z +3z + z 3 + z 4. y[n] =3δ[n ] + 3δ[n ] + δ[n 3] + δ[n 4]. Fact. Time-domain convolution of finite-duration signals is equivalent to z domain polynomial multiplication! Indeed, in MATLAB, one uses the conv command for multiplying polynomials! Proof of convolution property { } Y (z) = Z{y[n]} = Z{x[n] h[n]} = Z h[k] x[n k] = h[k] Z{x[n k]} by linearity of Z{ } k k = k h[k] X(z) z k by time-shift property of z-transform = X(z) k h[k] z k = X(z) H(z). Three laws of LTI systems (cf. Ohm s law) Given any two of the three {X(z),H(z),Y(z)} we can find the third as follows. All three are useful for different purposes. Y (z) =H(z) X(z) H(z) =Y (z) /X(z) X(z) =Y (z) /H(z)
6 c J. Fessler, June 9, 3, 6:3 (student version) 7.6 Finding H(z) from I/O relation We canuse therelationship H(z) =Y (z) /X(z) to find H(z) easily. Example. (Analysis) A -point moving average filter has the input-output relationship so in the z domain, using the linearity and time-shift properties: y[n] = x[n]+ x[n ] Y (z) = X(z)+ X(z) z = Dividing both sides by X(z) yields H(z) = Y (z) X(z) = + z. This approach is often the easiest way to find H(z), particularly for IIR filters. ( + ) z X(z). Pole-zero plots Example. Continuing the previous example, notice that H(z) = + z = z +. z The / is called the gain of the filter. The roots of the numerator polynomial are called the zeros of the system. In this example the only zero is at z =. The roots of the denominator polynomial are called the zeros of the system. In this example the only pole is at z =. We display the poles and zeros graphically using a pole-zero plot. gain =/ More generally, hereafter we will focus on rational system functions which are ratios of polynomials H(z) = B(z) A(z) = g i(z z i) j (z p j) = g (z z )(z z ) (z z M ) (z p )(z p ) (z p N ). B(z) is a Mth-order polynomial with M roots {z i,i=,...,m} called the zeros of H(z), because H(z i )=. A(z) is a Nth-order polynomial with N roots {p j,j =,...,N} called the poles of H(z), because H(p j ) =. Hereafter, often we will describe filters using the pole-zero plot as the graphical representation. The pole-zero plot (and gain) uniquely describe any rational system function. (New, and final, representation!) Example. A filter has the following pole-zero plot. Determine h[n]. gain =3 H(z) =3 (z j)(z + j) z =3 z + z =3+3z. Thus, by coefficient matching, h[n] =3δ[n]+3δ[n ].
7 c J. Fessler, June 9, 3, 6:3 (student version) 7.7 System functions of causal FIR filters Why were there poles only at the origin in the preceding two examples? Because FIR filters only have poles at the origin. Rewriting in two forms: M M h[n] = b k δ[n k] H(z) = b k z k = b + b z + + b M z M. k= k= H(z) = b z M + b z M + + b M z + b M (z z )(z z ) (z z M ) z M = b z M, where the z k s denote the M roots of the numerator polynomial, i.e.,them zeros of the system. Use MATLAB s roots function. Fact. A Mth-order causal FIR filter with b and b M,hasM zeros and has M poles at z =. We consider only the usual case of real filters (so the filter coefficients, the b k s, are real numbers). Then by the Fundamental theorem of algebra, all of the roots of the numerator polynomial are either real, or come in complex conjugate pairs. Fact. The zeros (and poles) of a real system are either real or come in complex conjugate pairs. Example. The system function for a FIR filter with impulse response h[n] =δ[n] δ[n ] + δ[n ] is: [ H(z) = z + z = z z + z z = ( + j )][ z ( j )] z, because the roots of the numerator polynomial Az + Bz + C are at z ± = B ± B 4AC B So the pole-zero diagram is: = ( ± j ). gain = The zeros are in complex-conjugate pairs. What about noncausal FIR filters? (Number of poles at the origin varies.) Examples. H(z) =z H(z) =z 3 3 For FIR filters, causal or not, any poles are at z =. poles at 3 zeros at
8 c J. Fessler, June 9, 3, 6:3 (student version) 7.8 The pole-zero plot and frequency response Why is the pole-zero plot useful? System design via frequency response! Relation between frequency response and system function: H(ˆω) =H(z) z=e j ˆω = H ( e j ˆω). So the magnitude response H(ˆω) = H ( e j ˆω ) From the general expression for H(z) above, the magnitude response of an FIR filter is H(ˆω) = ( H e j ˆω ) (e j ˆω z )(e j ˆω z ) (e j ˆω z M ) = b (e j ˆω ) M = b e j ˆω z e j ˆω z e j ˆω z M. So the magnitude response is simply the product of the distances from the point e j ˆω (on the unit circle) to all the zeros. In particular, whenever e j ˆω is close to a zero of the system function, the magnitude response will be reduced. And if e j ˆω = z m for one of the M zeros, then the magnitude response will be exactly zero. This makes it fairly easy to design filters that remove certain frequency components completely. And, given a pole-zero plot, we can roughly sketch the magnitude response! z =e j ˆω Ψˆω z =e jπ/4 z =e j The unit circle in the z-domain corresponds to the interval π ˆω π for the frequency domain. Example. A filter has the following pole-zero plot. Sketch its magnitude response. gain =3 For ˆω =, the distance to each zero is, so the magnitude response is 3 =6. Likewise for ˆω = ±π. As ˆω increases from towards π/, the distance to the upper zero decreases and the magnitude response drops towards zero. 6 H(ˆω) =6 cos ˆω π π/ π - ˆω Analytically: H(z) =3+3z so H(ˆω) =3+3e j ˆω =e j ˆω 6cosˆω, so the magnitude response is H(ˆω) =6 cos ˆω. For what sampling rate f s would this filter remove 6Hz hum? f s = 4Hz The book also discusses how to analyze the phase response graphically. This topic will be considered in more detail in EECS 45. 6
9 c J. Fessler, June 9, 3, 6:3 (student version) 7.9 First FIR filter design attempt: 6Hz notch filter. Equipped with the concepts developed thus far, we can finally attempt our first filter design! In many applications, such as measuring brain signals, 6Hz hum from AC power lines contaminates the signal x(t). Another example is home computer networks that operate over house AC wiring. To reduce the resulting interference, often we need to pass the signal through a filter that removes the 6Hz component while leaving other frequency components relatively unaffected. Such filtering can be done using RLC circuits, or by applying digital signal processing. x(t) Sample f s Filter H(ˆω) D/A (interpolator) y(t) x(t). 6Hz hum For simplicity of the algebra, we suppose that the sampling rate is f s = 48Hz. What is the corresponding digital frequency? ˆω =πf/f s =π6/48 = π/4. What is the ideal frequency response H(ˆω)? (Picture) of ideal H(ˆω) with nulls at ±π/4. This is called a notch filter. Design Where do we put zeros in the pole-zero diagram to eliminate the frequency component ˆω = ±π/4? See pole-zero diagram below with two zeros at ±e jπ/4. Where do the poles go? Two at origin needed for causal FIR filter. This system design looks promising, since it will certainly eliminate the 6Hz (ˆω = ±π/4) frequency component completely. Next we find the impulse response (needed for implementation) and the frequency response (to see if we met our design goal). Choosing gain= for now, we see from the pole-zero plot: [ ][ z e jπ/4 z e jπ/4] H(z) = z = z zcos(π/4) + z = z cos(π/4) + z = z +z. Thus the impulse response is: h[n] =δ[n] δ[n ] + δ[n ]. This is a very simple filter! It can be implemented using the following simple diffeq: y[n] =x[n] x[n ] + x[n ]. The frequency response of this FIR filter is H(ˆω) = e j ˆω +e j ˆω =e j ˆω ( cos ˆω ). Here is a plot of the magnitude response and phase response of this system, computed using MATLAB s zplane and freqz functions. Imaginary part.5.5 zplane Real part H(ω) Magnitude Response 4 3 π π/ ω π/ π Phase Response 5 h[n] H(ω) 5 5 π π/ ω π/ n π
10 c J. Fessler, June 9, 3, 6:3 (student version) 7. Design Our first design does remove the ˆω = ±π/4 frequency component completely, but unfortunately it also has the side effect of amplifying the high frequencies relative to the low frequencies. How can we attenuate the high frequencies somewhat? From the general expression for H(z) above, the magnitude response is H(ˆω) = ( H e j ˆω ) (e j ˆω z )(e j ˆω z ) (e j ˆω z M ) = (e j ˆω ) M = ej ˆω z e j ˆω z e j ˆω z M. So the magnitude response is simply the product of the distances from the point e j ˆω (on the unit circle) to all the zeros. In particular, whenever e j ˆω is close to a zero of the system function, the magnitude response will be reduced. So to reduce (but not eliminate) high frequency components (ˆω ±π), we can place a zero somewhere near z =e jπ =. Comparing H() = H() to H(π) =H( ), one can show that a zero at z = will equalize the gain at ˆω =and ˆω = π. Here is the resulting frequency response. Imaginary part.5.5 h[n] zplane 3 Real part H(ω) H(ω) Magnitude Response 4 3 π π/ ω π/ π Phase Response 5 5 n 5 π π/ ω π/ π Well, we have equalized the gain at ˆω =and ˆω = π, but we are still far from our ideal frequency response! What is the output of this filter if the input signal is x[n] =3+cos( π 4 n)+7( )n? Recall that x[n] =cos(ˆωn + φ) H(ˆω) y[n] = H(ˆω) cos(ˆωn + φ + H(ˆω)). For the system above, H(z) = [ z e jπ/4 ][ z e jπ/4][ z + ] z 3 H(ˆω) = H ( e j ˆω) =[ e j ˆω +e j ˆω ][ + e j ˆω ]. =[ z +z ][ + z ]= z + z 3 The frequencies of the input signal x[n] above are ˆω =, ˆω = π/4,andˆω = π. Substituting into H(ˆω) we have H() = ( )( + ) =, H(π/4) = (due to zeros at ±π/4 on unit circle), H(π) = ( + )( ) = = e jπ. Thus the output signal is y[n] =3 ++ cos(πn + π).
11 c J. Fessler, June 9, 3, 6:3 (student version) 7. A magnitude response plot shows the response to an eternal sinusoid. What about suddenly turned on signals cos(ˆωn)u[n], (e.g., after system first started)? After short initial transient response, the output signal approaches the steady state response. Design 3 How well can we do with a more sophisticated FIR filter? Here is a M =8filter designed by MATLAB s remez function. The frequency response plots were made using MATLAB s freqz function. The impulse response plots were made using MATLAB s filter function. Imaginary part.5.5 h[n] zplane 8 Real part n H(ω) H(ω) Magnitude Response.5.5 π π/ ω π/ π Phase Response 5 5 π π/ ω π/ π This is better since it has reasonably uniform gain in the passbands, but the notch may be somewhat wide. Design 4 How can we get even closer to the ideal frequency response? One way is to use the filter with the following pole-zero plot and frequency response. Imaginary part h[n] zplane Real part 5 n H(ω) H(ω) Magnitude Response.5.5 π π/ ω π/ π Phase Response π π/ ω π/ π
12 c J. Fessler, June 9, 3, 6:3 (student version) 7. What is fundamentally different about this filter? It is an IIR filter, since there are poles not at z =. Even though we have not yet studied IIR filters in detail, we can use what we have learned so far to find the system function H(z), the diffeq, and the frequency response. However, we do not yet know how to find the impulse response of an IIR filter. That comes in the next chapter. Assuming that the gain=, from the pole-zero we have that the system function is: H(z) = [z ejπ/4 ][z e jπ/4 ] [z re jπ/4 ][z re jπ/4 ] = z zcos(π/4) + z zr cos(π/4) + r = where r denotes the distance of the pole from the origin. To find the diffeq, wefirstcross multiply as follows: z cos(π/4) + z Y (z) z r cos(π/4) + r = z X(z), Y (z) [ z r cos(π/4) + r z ] = X(z) [ z cos(π/4) + z ]. Now (using linearity and shift property of z-transform), we convert back to the time domain: y[n] r cos(π/4)y[n ] + r y[n ] = x[n] cos(π/4)x[n ] + x[n ]. Rearranging yields the following recursive expression: y[n] =r cos(π/4)y[n ] r y[n ] + x[n] cos(π/4)x[n ] + x[n ]. The output at time n depends on the two past output values and on the current input and past input values. This is very easy to implement! Much easier than a 8-tap FIR filter. To understand the frequency response qualitatively, we return to the first factored form of the system function above: H(ˆω) =H ( e j ˆω) = [ej ˆω e jπ/4 ][e j ˆω e jπ/4 ] [e j ˆω re jπ/4 ][e j ˆω re jπ/4 ]. In particular, we find the magnitude response by taking the magnitude of both sides: H(ˆω) = ej ˆω e jπ/4 e j ˆω e jπ/4 e j ˆω re jπ/4 e j ˆω re jπ/4. In words, this means that for any frequency ˆω, the magnitude response is the product of all the distances from the point e j ˆω on the unit circle to all the zeros, divided by the product of all the distances from the point e j ˆω to all the poles. Being close to a zero decreases the magnitude response; being close to pole increases the magnitude response. For any frequency ˆω that is far from e jπ/4 in the notch filter example, the distance from e j ˆω to the zero and to its neighboring pole is almost the same, so the ratio is nearly unity and thus the magnitude response is nearly unity. The zero is slightly further away, so the magnitude response is slightly more than unity. How could we make the response closer to unity far from the zero? Move the pole closer to the zero. (But, as we will see in the next chapter, this makes the transient response longer, i.e., the impulse response decays to zero slower, which can be undesirable in some applications.
13 c J. Fessler, June 9, 3, 6:3 (student version) 7.3 The following figure shows the result of applying this filter to the example data considered much earlier. The transient response of the IIR filter is clearly visible. Input signal Input spectrum x[n] 5 X[k] Output nsignal 5 Output spectrum k y[n] 5 IIR filter, r=.85 b = [..4.] a = [...7] Y[k] n 5 k
14 c J. Fessler, June 9, 3, 6:3 (student version) 7.4 Interconnection of LTI systems Engineers build complicated systems by connecting simpler components. There are three principal ways of connecting two LTI systems: series, parallel, and in a feedback loop. Series connection Here are two LTI systems connected in series or in cascade: x[n] H (z) y[n] H (z) y[n]. In the time domain, we have seen that that overall system is LTI with impulse response h[n] =h [n] h [n]. Since time-domain convolution corresponds to multiplication in the z-domain, the overall system function is H(z) =H (z) H (z), the product of the two individual system functions. Example. Application: channel equalization for wireless communications. H (z) represents the channel (attenuation, multipath) and H (z) represents the equalizer that is designed to undo (invert) the distortions induced by the channel. Ideally, we would have H =/H. We must estimate the channel response H. We sent a training sequence with known x[n], observe the output y [n], and then use H (z) =Y (z) /X(z). These ideas are used in digital communications. Example. Application: (greatly simplified) Dolby noise reduction... gain= gain= High-boost for record: High-reduce for playback (inverse system: IIR!): (Picture) of frequency responses. Example. Putting two unit-sample delays in series is equivalent to a two-sample delay: x[n] z z y[n] is equivalent to x[n] z y[n]. Example. H (z) =g ( + z z+ )=g z In series: Pole-zero plots: and H (z) =g ( z )=g z z. z +z (z +)(z ) H(z) =H (z) H (z) =g g = g g z z z. H (z): For rational system functions: H (z): H(z) =H (z) H (z) = B (z) B (z) A (z) A (z). So the zeros of H(z) are all the zeros of the two systems connected in zeros, and likewise for the poles, except in cases where B (z) and A (z) share a common root, or where B (z) and A (z) share a common root, which is called pole-zero cancellation. H:
15 c J. Fessler, June 9, 3, 6:3 (student version) 7.5 Parallel connection x[n] - - H (z) H (z)? y[n] Y (z) =H (z) X(z)+H (z) X(z) =(H (z)+h (z))x(z), so the overall system function is H(z) =H (z)+h (z) = B (z) A (z) + B (z) A (z) = B (z) A (z)+b (z) A (z). A (z) A (z) In this case we see that the parallel connected system has all the poles of the two component systems (unless there is pole-zero cancellation. But we can say very little if anything about the zeros in general. Example. H (z) =g ( + z z+ )=g z and H (z) =g ( z z )=g z. In parallel: z + z H(z) =H (z)+h (z) =g + g = (g + g )z +(g g ). z z z Single pole at z =and zero at z =(g g )/(g + g ). Pole-zero plots: H (z): H (z): H (for g =, g =): Feedback connection Example. automobile cruise control. (Picture of system H (z) with negative feedback H f (z)) Derivation: Y = H (X H f Y ) so ( + H f )Y = H X so H(z) = H (z) +H (z) H f (z). Location of poles and zeros of H(z) is complicated; indeed, part of the purpose of feedback is to stabilize and otherwise unstable system (e.g., Harriet airplane). Even if H (z) has poles that are outside the unit circle, which means it is an unstable system, it can still be the case that H(z) has all its poles inside the unit circle, meaning that it is stable, if the feedback system H f (z) is designed properly. Example. H (z) = az = z z a. If a>, then the system is unstable. The simplest possible feedback controller would be H f (z) =g for some constant g. z H (z) H(z) = +H (z) H f (z) = z a + z z a g = z z a + zg = z z( + g) a, so the system with feedback has a pole at z = a/( + g). If g>a, then the pole will be moved to within the unit circle, thereby stabilizing the system.
16 c J. Fessler, June 9, 3, 6:3 (student version) 7.6 Example. Noise cancellation. (Explained in lecture) x[n] mic + + y[n] H (z) w[n] H (z) mic H 3 (z) Environment For perfect noise cancellation, we need H 3 (z) = H (z) /H (z). If H (z) is FIR, then H 3 (z) will be IIR, which again brings us to the next chapter...
17 c J. Fessler, June 9, 3, 6:3 (student version) 7.7 Summary and other topics system function H(z): z-transformof impulseresponse. pole-zero plot / filter design. Response to suddenly applied complex exponential signals and sinusoids? Design of FIR bandpass filter? (zeros around unit circle except in passband) (Picture) of ideal H(ˆω) (Picture) pole-zero plot
Lecture 19 IIR Filters
Lecture 19 IIR Filters Fundamentals of Digital Signal Processing Spring, 2012 Wei-Ta Chu 2012/5/10 1 General IIR Difference Equation IIR system: infinite-impulse response system The most general class
More informationz-transforms Definition of the z-transform Chapter
z-transforms Chapter 7 In the study of discrete-time signal and systems, we have thus far considered the time-domain and the frequency domain. The z- domain gives us a third representation. All three domains
More informationLet H(z) = P(z)/Q(z) be the system function of a rational form. Let us represent both P(z) and Q(z) as polynomials of z (not z -1 )
Review: Poles and Zeros of Fractional Form Let H() = P()/Q() be the system function of a rational form. Let us represent both P() and Q() as polynomials of (not - ) Then Poles: the roots of Q()=0 Zeros:
More informationVery useful for designing and analyzing signal processing systems
z-transform z-transform The z-transform generalizes the Discrete-Time Fourier Transform (DTFT) for analyzing infinite-length signals and systems Very useful for designing and analyzing signal processing
More informationDiscrete-time signals and systems
Discrete-time signals and systems 1 DISCRETE-TIME DYNAMICAL SYSTEMS x(t) G y(t) Linear system: Output y(n) is a linear function of the inputs sequence: y(n) = k= h(k)x(n k) h(k): impulse response of the
More informationSignals and Systems. Problem Set: The z-transform and DT Fourier Transform
Signals and Systems Problem Set: The z-transform and DT Fourier Transform Updated: October 9, 7 Problem Set Problem - Transfer functions in MATLAB A discrete-time, causal LTI system is described by the
More information# FIR. [ ] = b k. # [ ]x[ n " k] [ ] = h k. x[ n] = Ae j" e j# ˆ n Complex exponential input. [ ]Ae j" e j ˆ. ˆ )Ae j# e j ˆ. y n. y n.
[ ] = h k M [ ] = b k x[ n " k] FIR k= M [ ]x[ n " k] convolution k= x[ n] = Ae j" e j ˆ n Complex exponential input [ ] = h k M % k= [ ]Ae j" e j ˆ % M = ' h[ k]e " j ˆ & k= k = H (" ˆ )Ae j e j ˆ ( )
More informationE : Lecture 1 Introduction
E85.2607: Lecture 1 Introduction 1 Administrivia 2 DSP review 3 Fun with Matlab E85.2607: Lecture 1 Introduction 2010-01-21 1 / 24 Course overview Advanced Digital Signal Theory Design, analysis, and implementation
More informationModule 4 : Laplace and Z Transform Problem Set 4
Module 4 : Laplace and Z Transform Problem Set 4 Problem 1 The input x(t) and output y(t) of a causal LTI system are related to the block diagram representation shown in the figure. (a) Determine a differential
More informationDigital Filters Ying Sun
Digital Filters Ying Sun Digital filters Finite impulse response (FIR filter: h[n] has a finite numbers of terms. Infinite impulse response (IIR filter: h[n] has infinite numbers of terms. Causal filter:
More informationHow to manipulate Frequencies in Discrete-time Domain? Two Main Approaches
How to manipulate Frequencies in Discrete-time Domain? Two Main Approaches Difference Equations (an LTI system) x[n]: input, y[n]: output That is, building a system that maes use of the current and previous
More informationCMPT 889: Lecture 5 Filters
CMPT 889: Lecture 5 Filters Tamara Smyth, tamaras@cs.sfu.ca School of Computing Science, Simon Fraser University October 7, 2009 1 Digital Filters Any medium through which a signal passes may be regarded
More informationDigital Filters. Linearity and Time Invariance. Linear Time-Invariant (LTI) Filters: CMPT 889: Lecture 5 Filters
Digital Filters CMPT 889: Lecture 5 Filters Tamara Smyth, tamaras@cs.sfu.ca School of Computing Science, Simon Fraser University October 7, 29 Any medium through which a signal passes may be regarded as
More informationDiscrete-Time Signals and Systems. Frequency Domain Analysis of LTI Systems. The Frequency Response Function. The Frequency Response Function
Discrete-Time Signals and s Frequency Domain Analysis of LTI s Dr. Deepa Kundur University of Toronto Reference: Sections 5., 5.2-5.5 of John G. Proakis and Dimitris G. Manolakis, Digital Signal Processing:
More information7.17. Determine the z-transform and ROC for the following time signals: Sketch the ROC, poles, and zeros in the z-plane. X(z) = x[n]z n.
Solutions to Additional Problems 7.7. Determine the -transform and ROC for the following time signals: Sketch the ROC, poles, and eros in the -plane. (a) x[n] δ[n k], k > 0 X() x[n] n n k, 0 Im k multiple
More informationz-transform Chapter 6
z-transform Chapter 6 Dr. Iyad djafar Outline 2 Definition Relation Between z-transform and DTFT Region of Convergence Common z-transform Pairs The Rational z-transform The Inverse z-transform z-transform
More informationDetailed Solutions to Exercises
Detailed Solutions to Exercises Digital Signal Processing Mikael Swartling Nedelko Grbic rev. 205 Department of Electrical and Information Technology Lund University Detailed solution to problem E3.4 A
More informationAnalog vs. discrete signals
Analog vs. discrete signals Continuous-time signals are also known as analog signals because their amplitude is analogous (i.e., proportional) to the physical quantity they represent. Discrete-time signals
More informationZ Transform (Part - II)
Z Transform (Part - II). The Z Transform of the following real exponential sequence x(nt) = a n, nt 0 = 0, nt < 0, a > 0 (a) ; z > (c) for all z z (b) ; z (d) ; z < a > a az az Soln. The given sequence
More informationResponses of Digital Filters Chapter Intended Learning Outcomes:
Responses of Digital Filters Chapter Intended Learning Outcomes: (i) Understanding the relationships between impulse response, frequency response, difference equation and transfer function in characterizing
More informationUse: Analysis of systems, simple convolution, shorthand for e jw, stability. Motivation easier to write. Or X(z) = Z {x(n)}
1 VI. Z Transform Ch 24 Use: Analysis of systems, simple convolution, shorthand for e jw, stability. A. Definition: X(z) = x(n) z z - transforms Motivation easier to write Or Note if X(z) = Z {x(n)} z
More informationUNIVERSITY OF OSLO. Please make sure that your copy of the problem set is complete before you attempt to answer anything.
UNIVERSITY OF OSLO Faculty of mathematics and natural sciences Examination in INF3470/4470 Digital signal processing Day of examination: December 9th, 011 Examination hours: 14.30 18.30 This problem set
More informationCosc 3451 Signals and Systems. What is a system? Systems Terminology and Properties of Systems
Cosc 3451 Signals and Systems Systems Terminology and Properties of Systems What is a system? an entity that manipulates one or more signals to yield new signals (often to accomplish a function) can be
More informationSignal Processing First Lab 11: PeZ - The z, n, and ˆω Domains
Signal Processing First Lab : PeZ - The z, n, and ˆω Domains The lab report/verification will be done by filling in the last page of this handout which addresses a list of observations to be made when
More informationAn Fir-Filter Example: Hanning Filter
An Fir-Filter Example: Hanning Filter Josef Goette Bern University of Applied Sciences, Biel Institute of Human Centered Engineering - microlab Josef.Goette@bfh.ch February 7, 2018 Contents 1 Mathematical
More informationVI. Z Transform and DT System Analysis
Summer 2008 Signals & Systems S.F. Hsieh VI. Z Transform and DT System Analysis Introduction Why Z transform? a DT counterpart of the Laplace transform in CT. Generalization of DT Fourier transform: z
More informationDiscrete Time Systems
Discrete Time Systems Valentina Hubeika, Jan Černocký DCGM FIT BUT Brno, {ihubeika,cernocky}@fit.vutbr.cz 1 LTI systems In this course, we work only with linear and time-invariant systems. We talked about
More informationConvolution. Define a mathematical operation on discrete-time signals called convolution, represented by *. Given two discrete-time signals x 1, x 2,
Filters Filters So far: Sound signals, connection to Fourier Series, Introduction to Fourier Series and Transforms, Introduction to the FFT Today Filters Filters: Keep part of the signal we are interested
More informationUNIT-II Z-TRANSFORM. This expression is also called a one sided z-transform. This non causal sequence produces positive powers of z in X (z).
Page no: 1 UNIT-II Z-TRANSFORM The Z-Transform The direct -transform, properties of the -transform, rational -transforms, inversion of the transform, analysis of linear time-invariant systems in the -
More informationUniversity of Illinois at Urbana-Champaign ECE 310: Digital Signal Processing
University of Illinois at Urbana-Champaign ECE 0: Digital Signal Processing Chandra Radhakrishnan PROBLEM SET : SOLUTIONS Peter Kairouz Problem. Hz z 7 z +/9, causal ROC z > contains the unit circle BIBO
More informationEE102B Signal Processing and Linear Systems II. Solutions to Problem Set Nine Spring Quarter
EE02B Signal Processing and Linear Systems II Solutions to Problem Set Nine 202-203 Spring Quarter Problem 9. (25 points) (a) 0.5( + 4z + 6z 2 + 4z 3 + z 4 ) + 0.2z 0.4z 2 + 0.8z 3 x[n] 0.5 y[n] -0.2 Z
More informationDiscrete-Time Signals and Systems. The z-transform and Its Application. The Direct z-transform. Region of Convergence. Reference: Sections
Discrete-Time Signals and Systems The z-transform and Its Application Dr. Deepa Kundur University of Toronto Reference: Sections 3. - 3.4 of John G. Proakis and Dimitris G. Manolakis, Digital Signal Processing:
More informationTransform analysis of LTI systems Oppenheim and Schafer, Second edition pp For LTI systems we can write
Transform analysis of LTI systems Oppenheim and Schafer, Second edition pp. 4 9. For LTI systems we can write yœn D xœn hœn D X kd xœkhœn Alternatively, this relationship can be expressed in the z-transform
More informationDigital Signal Processing
COMP ENG 4TL4: Digital Signal Processing Notes for Lecture #21 Friday, October 24, 2003 Types of causal FIR (generalized) linear-phase filters: Type I: Symmetric impulse response: with order M an even
More informationECE503: Digital Signal Processing Lecture 5
ECE53: Digital Signal Processing Lecture 5 D. Richard Brown III WPI 3-February-22 WPI D. Richard Brown III 3-February-22 / 32 Lecture 5 Topics. Magnitude and phase characterization of transfer functions
More informationModule 1: Signals & System
Module 1: Signals & System Lecture 6: Basic Signals in Detail Basic Signals in detail We now introduce formally some of the basic signals namely 1) The Unit Impulse function. 2) The Unit Step function
More informationStability Condition in Terms of the Pole Locations
Stability Condition in Terms of the Pole Locations A causal LTI digital filter is BIBO stable if and only if its impulse response h[n] is absolutely summable, i.e., 1 = S h [ n] < n= We now develop a stability
More informationIntroduction to DSP Time Domain Representation of Signals and Systems
Introduction to DSP Time Domain Representation of Signals and Systems Dr. Waleed Al-Hanafy waleed alhanafy@yahoo.com Faculty of Electronic Engineering, Menoufia Univ., Egypt Digital Signal Processing (ECE407)
More informationPS403 - Digital Signal processing
PS403 - Digital Signal processing 6. DSP - Recursive (IIR) Digital Filters Key Text: Digital Signal Processing with Computer Applications (2 nd Ed.) Paul A Lynn and Wolfgang Fuerst, (Publisher: John Wiley
More informationEEL3135: Homework #4
EEL335: Homework #4 Problem : For each of the systems below, determine whether or not the system is () linear, () time-invariant, and (3) causal: (a) (b) (c) xn [ ] cos( 04πn) (d) xn [ ] xn [ ] xn [ 5]
More informationDigital Signal Processing Lecture 4
Remote Sensing Laboratory Dept. of Information Engineering and Computer Science University of Trento Via Sommarive, 14, I-38123 Povo, Trento, Italy Digital Signal Processing Lecture 4 Begüm Demir E-mail:
More informationEE 3054: Signals, Systems, and Transforms Spring A causal discrete-time LTI system is described by the equation. y(n) = 1 4.
EE : Signals, Systems, and Transforms Spring 7. A causal discrete-time LTI system is described by the equation Test y(n) = X x(n k) k= No notes, closed book. Show your work. Simplify your answers.. A discrete-time
More informationFourier Series Representation of
Fourier Series Representation of Periodic Signals Rui Wang, Assistant professor Dept. of Information and Communication Tongji University it Email: ruiwang@tongji.edu.cn Outline The response of LIT system
More informationChap 2. Discrete-Time Signals and Systems
Digital Signal Processing Chap 2. Discrete-Time Signals and Systems Chang-Su Kim Discrete-Time Signals CT Signal DT Signal Representation 0 4 1 1 1 2 3 Functional representation 1, n 1,3 x[ n] 4, n 2 0,
More informationAnalog LTI system Digital LTI system
Sampling Decimation Seismometer Amplifier AAA filter DAA filter Analog LTI system Digital LTI system Filtering (Digital Systems) input output filter xn [ ] X ~ [ k] Convolution of Sequences hn [ ] yn [
More informationECGR4124 Digital Signal Processing Exam 2 Spring 2017
ECGR4124 Digital Signal Processing Exam 2 Spring 2017 Name: LAST 4 NUMBERS of Student Number: Do NOT begin until told to do so Make sure that you have all pages before starting NO TEXTBOOK, NO CALCULATOR,
More informationRecursive, Infinite Impulse Response (IIR) Digital Filters:
Recursive, Infinite Impulse Response (IIR) Digital Filters: Filters defined by Laplace Domain transfer functions (analog devices) can be easily converted to Z domain transfer functions (digital, sampled
More informationDigital Filter Structures. Basic IIR Digital Filter Structures. of an LTI digital filter is given by the convolution sum or, by the linear constant
Digital Filter Chapter 8 Digital Filter Block Diagram Representation Equivalent Basic FIR Digital Filter Basic IIR Digital Filter. Block Diagram Representation In the time domain, the input-output relations
More informationDiscrete Time Systems
1 Discrete Time Systems {x[0], x[1], x[2], } H {y[0], y[1], y[2], } Example: y[n] = 2x[n] + 3x[n-1] + 4x[n-2] 2 FIR and IIR Systems FIR: Finite Impulse Response -- non-recursive y[n] = 2x[n] + 3x[n-1]
More informationELEG 305: Digital Signal Processing
ELEG 305: Digital Signal Processing Lecture 4: Inverse z Transforms & z Domain Analysis Kenneth E. Barner Department of Electrical and Computer Engineering University of Delaware Fall 008 K. E. Barner
More informationZ-TRANSFORMS. Solution: Using the definition (5.1.2), we find: for case (b). y(n)= h(n) x(n) Y(z)= H(z)X(z) (convolution) (5.1.
84 5. Z-TRANSFORMS 5 z-transforms Solution: Using the definition (5..2), we find: for case (a), and H(z) h 0 + h z + h 2 z 2 + h 3 z 3 2 + 3z + 5z 2 + 2z 3 H(z) h 0 + h z + h 2 z 2 + h 3 z 3 + h 4 z 4
More informationChapter Intended Learning Outcomes: (i) Understanding the relationship between transform and the Fourier transform for discrete-time signals
z Transform Chapter Intended Learning Outcomes: (i) Understanding the relationship between transform and the Fourier transform for discrete-time signals (ii) Understanding the characteristics and properties
More informationLecture 11 FIR Filters
Lecture 11 FIR Filters Fundamentals of Digital Signal Processing Spring, 2012 Wei-Ta Chu 2012/4/12 1 The Unit Impulse Sequence Any sequence can be represented in this way. The equation is true if k ranges
More informationDSP First. Laboratory Exercise #10. The z, n, and ˆω Domains
DSP First Laboratory Exercise #10 The z, n, and ˆω Domains 1 Objective The objective for this lab is to build an intuitive understanding of the relationship between the location of poles and zeros in the
More information(i) Represent discrete-time signals using transform. (ii) Understand the relationship between transform and discrete-time Fourier transform
z Transform Chapter Intended Learning Outcomes: (i) Represent discrete-time signals using transform (ii) Understand the relationship between transform and discrete-time Fourier transform (iii) Understand
More informationLecture 2 OKAN UNIVERSITY FACULTY OF ENGINEERING AND ARCHITECTURE
OKAN UNIVERSITY FACULTY OF ENGINEERING AND ARCHITECTURE EEE 43 DIGITAL SIGNAL PROCESSING (DSP) 2 DIFFERENCE EQUATIONS AND THE Z- TRANSFORM FALL 22 Yrd. Doç. Dr. Didem Kivanc Tureli didemk@ieee.org didem.kivanc@okan.edu.tr
More informationExamples. 2-input, 1-output discrete-time systems: 1-input, 1-output discrete-time systems:
Discrete-Time s - I Time-Domain Representation CHAPTER 4 These lecture slides are based on "Digital Signal Processing: A Computer-Based Approach, 4th ed." textbook by S.K. Mitra and its instructor materials.
More information2. Typical Discrete-Time Systems All-Pass Systems (5.5) 2.2. Minimum-Phase Systems (5.6) 2.3. Generalized Linear-Phase Systems (5.
. Typical Discrete-Time Systems.1. All-Pass Systems (5.5).. Minimum-Phase Systems (5.6).3. Generalized Linear-Phase Systems (5.7) .1. All-Pass Systems An all-pass system is defined as a system which has
More informationSolutions. Number of Problems: 10
Final Exam February 2nd, 2013 Signals & Systems (151-0575-01) Prof. R. D Andrea Solutions Exam Duration: 150 minutes Number of Problems: 10 Permitted aids: One double-sided A4 sheet. Questions can be answered
More informationECE503: Digital Signal Processing Lecture 4
ECE503: Digital Signal Processing Lecture 4 D. Richard Brown III WPI 06-February-2012 WPI D. Richard Brown III 06-February-2012 1 / 29 Lecture 4 Topics 1. Motivation for the z-transform. 2. Definition
More informationy[n] = = h[k]x[n k] h[k]z n k k= 0 h[k]z k ) = H(z)z n h[k]z h (7.1)
7. The Z-transform 7. Definition of the Z-transform We saw earlier that complex exponential of the from {e jwn } is an eigen function of for a LTI System. We can generalize this for signals of the form
More informationReview of Discrete-Time System
Review of Discrete-Time System Electrical & Computer Engineering University of Maryland, College Park Acknowledgment: ENEE630 slides were based on class notes developed by Profs. K.J. Ray Liu and Min Wu.
More informationDiscrete-Time Systems
FIR Filters With this chapter we turn to systems as opposed to signals. The systems discussed in this chapter are finite impulse response (FIR) digital filters. The term digital filter arises because these
More informationLike bilateral Laplace transforms, ROC must be used to determine a unique inverse z-transform.
Inversion of the z-transform Focus on rational z-transform of z 1. Apply partial fraction expansion. Like bilateral Laplace transforms, ROC must be used to determine a unique inverse z-transform. Let X(z)
More informationZ-Transform. x (n) Sampler
Chapter Two A- Discrete Time Signals: The discrete time signal x(n) is obtained by taking samples of the analog signal xa (t) every Ts seconds as shown in Figure below. Analog signal Discrete time signal
More informationIII. Time Domain Analysis of systems
1 III. Time Domain Analysis of systems Here, we adapt properties of continuous time systems to discrete time systems Section 2.2-2.5, pp 17-39 System Notation y(n) = T[ x(n) ] A. Types of Systems Memoryless
More informationFinal Exam January 31, Solutions
Final Exam January 31, 014 Signals & Systems (151-0575-01) Prof. R. D Andrea & P. Reist Solutions Exam Duration: Number of Problems: Total Points: Permitted aids: Important: 150 minutes 7 problems 50 points
More informationDigital Signal Processing:
Digital Signal Processing: Mathematical and algorithmic manipulation of discretized and quantized or naturally digital signals in order to extract the most relevant and pertinent information that is carried
More informationZ-Transform. 清大電機系林嘉文 Original PowerPoint slides prepared by S. K. Mitra 4-1-1
Chapter 6 Z-Transform 清大電機系林嘉文 cwlin@ee.nthu.edu.tw 03-5731152 Original PowerPoint slides prepared by S. K. Mitra 4-1-1 z-transform The DTFT provides a frequency-domain representation of discrete-time
More information( ) John A. Quinn Lecture. ESE 531: Digital Signal Processing. Lecture Outline. Frequency Response of LTI System. Example: Zero on Real Axis
John A. Quinn Lecture ESE 531: Digital Signal Processing Lec 15: March 21, 2017 Review, Generalized Linear Phase Systems Penn ESE 531 Spring 2017 Khanna Lecture Outline!!! 2 Frequency Response of LTI System
More informationUNIVERSITY OF OSLO. Faculty of mathematics and natural sciences. Forslag til fasit, versjon-01: Problem 1 Signals and systems.
UNIVERSITY OF OSLO Faculty of mathematics and natural sciences Examination in INF3470/4470 Digital signal processing Day of examination: December 1th, 016 Examination hours: 14:30 18.30 This problem set
More informationAPPLIED SIGNAL PROCESSING
APPLIED SIGNAL PROCESSING DIGITAL FILTERS Digital filters are discrete-time linear systems { x[n] } G { y[n] } Impulse response: y[n] = h[0]x[n] + h[1]x[n 1] + 2 DIGITAL FILTER TYPES FIR (Finite Impulse
More informationDiscrete-Time Fourier Transform (DTFT)
Discrete-Time Fourier Transform (DTFT) 1 Preliminaries Definition: The Discrete-Time Fourier Transform (DTFT) of a signal x[n] is defined to be X(e jω ) x[n]e jωn. (1) In other words, the DTFT of x[n]
More informationLinear Convolution Using FFT
Linear Convolution Using FFT Another useful property is that we can perform circular convolution and see how many points remain the same as those of linear convolution. When P < L and an L-point circular
More informationINFINITE-IMPULSE RESPONSE DIGITAL FILTERS Classical analog filters and their conversion to digital filters 4. THE BUTTERWORTH ANALOG FILTER
INFINITE-IMPULSE RESPONSE DIGITAL FILTERS Classical analog filters and their conversion to digital filters. INTRODUCTION 2. IIR FILTER DESIGN 3. ANALOG FILTERS 4. THE BUTTERWORTH ANALOG FILTER 5. THE CHEBYSHEV-I
More informationELEG 305: Digital Signal Processing
ELEG 305: Digital Signal Processing Lecture : Design of Digital IIR Filters (Part I) Kenneth E. Barner Department of Electrical and Computer Engineering University of Delaware Fall 008 K. E. Barner (Univ.
More informationThe Discrete-Time Fourier
Chapter 3 The Discrete-Time Fourier Transform 清大電機系林嘉文 cwlin@ee.nthu.edu.tw 03-5731152 Original PowerPoint slides prepared by S. K. Mitra 3-1-1 Continuous-Time Fourier Transform Definition The CTFT of
More informationFilter structures ELEC-E5410
Filter structures ELEC-E5410 Contents FIR filter basics Ideal impulse responses Polyphase decomposition Fractional delay by polyphase structure Nyquist filters Half-band filters Gibbs phenomenon Discrete-time
More information21.4. Engineering Applications of z-transforms. Introduction. Prerequisites. Learning Outcomes
Engineering Applications of z-transforms 21.4 Introduction In this Section we shall apply the basic theory of z-transforms to help us to obtain the response or output sequence for a discrete system. This
More informationMultirate signal processing
Multirate signal processing Discrete-time systems with different sampling rates at various parts of the system are called multirate systems. The need for such systems arises in many applications, including
More informationZ-Transform. The Z-transform is the Discrete-Time counterpart of the Laplace Transform. Laplace : G(s) = g(t)e st dt. Z : G(z) =
Z-Transform The Z-transform is the Discrete-Time counterpart of the Laplace Transform. Laplace : G(s) = Z : G(z) = It is Used in Digital Signal Processing n= g(t)e st dt g[n]z n Used to Define Frequency
More informationTransform Analysis of Linear Time-Invariant Systems
Transform Analysis of Linear Time-Invariant Systems Discrete-Time Signal Processing Chia-Ping Chen Department of Computer Science and Engineering National Sun Yat-Sen University Kaohsiung, Taiwan ROC Transform
More informationECE4270 Fundamentals of DSP Lecture 20. Fixed-Point Arithmetic in FIR and IIR Filters (part I) Overview of Lecture. Overflow. FIR Digital Filter
ECE4270 Fundamentals of DSP Lecture 20 Fixed-Point Arithmetic in FIR and IIR Filters (part I) School of ECE Center for Signal and Information Processing Georgia Institute of Technology Overview of Lecture
More informationECE 410 DIGITAL SIGNAL PROCESSING D. Munson University of Illinois Chapter 12
. ECE 40 DIGITAL SIGNAL PROCESSING D. Munson University of Illinois Chapter IIR Filter Design ) Based on Analog Prototype a) Impulse invariant design b) Bilinear transformation ( ) ~ widely used ) Computer-Aided
More informationECE538 Final Exam Fall 2017 Digital Signal Processing I 14 December Cover Sheet
ECE58 Final Exam Fall 7 Digital Signal Processing I December 7 Cover Sheet Test Duration: hours. Open Book but Closed Notes. Three double-sided 8.5 x crib sheets allowed This test contains five problems.
More informationLecture 04: Discrete Frequency Domain Analysis (z-transform)
Lecture 04: Discrete Frequency Domain Analysis (z-transform) John Chiverton School of Information Technology Mae Fah Luang University 1st Semester 2009/ 2552 Outline Overview Lecture Contents Introduction
More informationLecture 5 - Assembly Programming(II), Intro to Digital Filters
GoBack Lecture 5 - Assembly Programming(II), Intro to Digital Filters James Barnes (James.Barnes@colostate.edu) Spring 2009 Colorado State University Dept of Electrical and Computer Engineering ECE423
More informationLecture 7 Discrete Systems
Lecture 7 Discrete Systems EE 52: Instrumentation and Measurements Lecture Notes Update on November, 29 Aly El-Osery, Electrical Engineering Dept., New Mexico Tech 7. Contents The z-transform 2 Linear
More informationDigital Signal Processing I Final Exam Fall 2008 ECE Dec Cover Sheet
Digital Signal Processing I Final Exam Fall 8 ECE538 7 Dec.. 8 Cover Sheet Test Duration: minutes. Open Book but Closed Notes. Calculators NOT allowed. This test contains FIVE problems. All work should
More informationDiscrete-Time Signals & Systems
Chapter 2 Discrete-Time Signals & Systems 清大電機系林嘉文 cwlin@ee.nthu.edu.tw 03-5731152 Original PowerPoint slides prepared by S. K. Mitra 2-1-1 Discrete-Time Signals: Time-Domain Representation (1/10) Signals
More informationLecture 7 - IIR Filters
Lecture 7 - IIR Filters James Barnes (James.Barnes@colostate.edu) Spring 204 Colorado State University Dept of Electrical and Computer Engineering ECE423 / 2 Outline. IIR Filter Representations Difference
More informationECE-314 Fall 2012 Review Questions for Midterm Examination II
ECE-314 Fall 2012 Review Questions for Midterm Examination II First, make sure you study all the problems and their solutions from homework sets 4-7. Then work on the following additional problems. Problem
More informationMultidimensional digital signal processing
PSfrag replacements Two-dimensional discrete signals N 1 A 2-D discrete signal (also N called a sequence or array) is a function 2 defined over thex(n set 1 of, n 2 ordered ) pairs of integers: y(nx 1,
More informationZ - Transform. It offers the techniques for digital filter design and frequency analysis of digital signals.
Z - Transform The z-transform is a very important tool in describing and analyzing digital systems. It offers the techniques for digital filter design and frequency analysis of digital signals. Definition
More informationUniversity Question Paper Solution
Unit 1: Introduction University Question Paper Solution 1. Determine whether the following systems are: i) Memoryless, ii) Stable iii) Causal iv) Linear and v) Time-invariant. i) y(n)= nx(n) ii) y(t)=
More informationIntroduction to Signals and Systems Lecture #4 - Input-output Representation of LTI Systems Guillaume Drion Academic year
Introduction to Signals and Systems Lecture #4 - Input-output Representation of LTI Systems Guillaume Drion Academic year 2017-2018 1 Outline Systems modeling: input/output approach of LTI systems. Convolution
More informationModule 4. Related web links and videos. 1. FT and ZT
Module 4 Laplace transforms, ROC, rational systems, Z transform, properties of LT and ZT, rational functions, system properties from ROC, inverse transforms Related web links and videos Sl no Web link
More informationDigital Filter Structures
Chapter 8 Digital Filter Structures 清大電機系林嘉文 cwlin@ee.nthu.edu.tw 03-5731152 8-1 Block Diagram Representation The convolution sum description of an LTI discrete-time system can, in principle, be used to
More informationLECTURE NOTES DIGITAL SIGNAL PROCESSING III B.TECH II SEMESTER (JNTUK R 13)
LECTURE NOTES ON DIGITAL SIGNAL PROCESSING III B.TECH II SEMESTER (JNTUK R 13) FACULTY : B.V.S.RENUKA DEVI (Asst.Prof) / Dr. K. SRINIVASA RAO (Assoc. Prof) DEPARTMENT OF ELECTRONICS AND COMMUNICATIONS
More informationLecture 28 Continuous-Time Fourier Transform 2
Lecture 28 Continuous-Time Fourier Transform 2 Fundamentals of Digital Signal Processing Spring, 2012 Wei-Ta Chu 2012/6/14 1 Limit of the Fourier Series Rewrite (11.9) and (11.10) as As, the fundamental
More information