2.161 Signal Processing: Continuous and Discrete

Transcription

1 MI OpenCourseWare Signal Processing: Continuous and Discrete Fall 8 For information about citing these materials or our erms of Use, visit:

2 MASSACHUSES INSIUE OF ECHNOLOGY DEPARMEN OF MECHANICAL ENGINEERING.6 Signal Processing - Continuous and Discrete Fall erm 8 Solution of Problem Set Assigned: Sept. 8, 8 Due: Sept. 5, 8 Problem : Y (jω) = H(jΩ)F(jΩ) and it is shown in the below figure Y ( jw) y(t) = π Y (jω)e jωt dt 4 = ( e jωt dt + e jωt dt) π 4 4jt jt 4jt = (e e e + e jt ) πjt = (sin(4t) sin(t)) πt It is not a casual filter (since y(t) = h(t)/5 and h(t) = for some t < ). Problem : We define H (jω) = H(jΩ). hen H (jω) is a low pas filter, matching Prob. 5 in PS, which we have already found it s impulse response: H(jΩ) = H (jω) F (H(jΩ)) = F () F (H (jω)) h(t) = δ(t) h (t) h(t) = δ(t) sin(ω ct) πt

3 Problem 3: x(t) = A n sin(nω t + φ n ) = A n (sin(nω t) cosφ n + cos(nω t) sin φ n ) n= n= X(jΩ) = An ( F(sin(nΩ t)) cosφ n + F(cos(nΩ t)) sin φ n ) n= X(jΩ) = An (cosφ n ( jπ (δ(ω nω ) δ(ω + nω ))) + sin φ n (π (δ(ω nω ) + δ(ω + nω )))) n= X(jΩ) = jπ An ((cos φ n + j sin φ n ) (δ(ω nω ) + ( cosφ n + j sin φ n )(δ(ω + nω )) n= ( X(jΩ) = jπ An e jφ n δ(ω nω ) e jφn δ(ω + nω ) ) n= Problem 4: (a) he ideal multiplicative filtering operation is a low pass filtering with the pass-band Ω c = NΩ : { n N, Ω Ωc H n = n > N, Ω > Ω c (b) It s a convolution in this specific form: X n = X n H n x(t) = x(t) h(t) = x(τ)h(t τ)dτ We can prove that why convolution is in this specific integral form for our Periodic Exponential Fourier ransform: ( ( ) ) ( Xn H n e ) jnω t jnω x(t) = = X τ dτ e jnω t n h(τ)e n= n= ( jnω )) ( ( )) t ( jnω jnω (t τ) Xn e τ x(t) = h(τ) e dτ = h(τ) Xn e dτ x(t) = n= h(τ)x(t τ)dτ We can also find the specific form of our filter: n= N N N jnω t jnω t h(t) = H n e = e = + cos (nω t) = + cos (nω t) n= n= N n= n=

4 Here we use below trigonometric relation to simplify our summation, where ϕ = and α = Ω t: ( ) (n+)α nα sin cos (ϕ + ) cos ϕ + cos (ϕ + α) + cos (ϕ + α) + + cos (ϕ + nα) = sin ( ) N (N+)Ω t sin cos ( NΩt ) h(t) = + cos (nω t) = + Ω sin t n= hen we use below trigonometric relation to simplify our impulse response: sin θ cosϕ = sin(θ + ϕ) + sin(θ ϕ) ( ) (N+)Ω t sin cos ( NΩt ( ) sin (N + )Ω t ) + sin ( Ωt ) h(t) = + = + Ω sin t Ω sin t sin ( (N + )Ω t ) h(t) = sin Ω t α Here we further analyze this h(t) function, so we can use its properties for the next part of the problem: Note that for small t values, h(t) can be approximated to: sin ( (N + )Ω t ) t h(t) sin (NΩ t) he value of h(t) t, for large N values, is very close to h = which can be πt obtained for a Continuous Fourier ransform of a Low Pass Filter with Ω c = NΩ. Hence, in the vicinity of t =, h(t) acts like a sinc function (with the maximum value of N + ), but as soon as it gets close to its boundaries ( t Ω t π), it oscillates quickly with Ω = (N +.5)Ω around and +. Note that for the Continuous Fourier ransform, we expect a h (t); where it is to be evaluated between to + and contained with an envelope in the form of. t On the other hand, for our case of Periodic Fourier ransform, h(t) it is to be evaluated between to + and is also periodic. Note that for any N value, if x(t) then x (t) which means that following relation holds for any N: = h(τ)dτ = Ω t h(τ)dτ Furthermore, for very large N values, the h(t) function become very narrow, and hence above integral can be approximated to the integration around any finite, non-zero interval around t = : (Eq. ) N t s.t. < t 3 t : h(τ)dτ h(τ)dτ

5 his properties can be used to prove that x (t) values converge to x(t) values at any continuous point and to the mean of right and left limits at any stepwise discontinuous point of x(t). Furthermore, the structure of h(t) shows that the ripple will vanish at any point in which x(t) is continuous. Besides, as we will see in the next part, only a finite and determined value of ripple will be allowed to remain and it will be pushed to the edge of discontinuities. (c) he output signal is a convolution of the original signal with the h(t). Hence, it is very clear that ripples are due to deviation of h(t) from ideal case of δ(t). Consequently, as soon as N goes up, h(t) becomes more like a δ(t) function, acts more locally, and the output signal ripples with an increases frequency. For very large N values, the ripple percentage is a fixed amount, although its location is dependent on N value. he amount of ripple is fixed, because although different h(t) functions have different curves, but always the maximum ripple corresponds to the point where the edge of the central lobe matches with the discontinuity. We will prove this rigorously, but in fact since the ratio of the area of the central lobe, to the rest of the lobes remains constant, the ripple percentage remains constant as well. Note that due to Eq., only the behavior of function around discontinuity matters (as long as discontinuities have a finite non-zero distance from each other). Hence, without the loss of generality, we extend discontinuity limits to the the full extent of the function and consider our function in the below form where A value corresponds to the discontinuity jump: { A t < x(t) = t < Below figure shows this conditions: A x(t), to the scale t_max h(t) for N=, not to the scale / / By moving h(t) on the x(t), we can realize that, at each point, the x (t) is an average of neighborhood points. Particularly, the sign of lobes determine that whether those 4

6 point have a positive or negative contribution. Since, major contribution comes from the central lobes, we can guess that the maximum ripple occurs when the edge of central lobe match with t =. In that case, the next negative lobe cannot decrease the x(t) value and we have a maximum ripple. he next maxima and minima of ripples also correspond to other lobe edges touching t =. Now we prove this rigorously. For our specific case of x(t), and our even h(t) function, we can simplify the convolution integral: A A x (t) = h(t τ)x(τ)dτ = h(t τ)dτ = t dx (t) = h( t) + h(t) dt db x(t) t A max t max x(t max ) = h(τ)dτ = A( h(τ)dτ + h(τ)dτ) t max t max t max h(τ)dτ) = h(τ)dτ) = t h(τ)dτ We are interested in < t <<, and hence h(t). Consequently, maxi dt mum/minimum values correspond to (N + )Ω t max/min = kπ such that < k << N. π he highest maximum corresponds to the t max = =. Now we compute (N+ )Ω (N+ ) the x (t max ) value by breaking the original integral to two parts: Now note that for large N values from Eq. we can conclude that: Also for the other part of integral, τ is very close to zero and we can simplify h(t) with a sinc form and also change integration variable by θ = (N + )Ω t: h(τ)dτ) = t max sin ((N + )Ω τ) sin θ π sin θ dτ = dθ = dθ τ π θ Ω θ t max A short survey of literature or a aylor expansion of Ω sin θ θ π sin θ dθ = π θ Hence the maximum ripple corresponds to: ends to: x (t max ) = A( ) = A( ) π his corresponds to about 9% overshoot and this overshoot is only dependent on the local discontinuity jump and independent of N or specific form of x(t). Other proofs, for less general cases, could also be found in the Wikis. 5

7 : H : : H H! G G! G Problem 5: (a) s 3 H(s) = K s + 3 jω 3 H(jΩ) = K jω + 3 = K (b) Consider this pole-zero plot: M I F = A I Let the vectors from the poles and zeros to an arbitrary test frequency Ω be r i and q i : q q q 3 H(jΩ) = K r r r = K since q i = r i for all i regardless of the system order. herefore, this given pole-zero configuration, as well as all who satisfy problem conditions, are all-pass filters. (c) MALAB Command line : >> H_s=tf([ -3],[ 3]) ransfer function: s s + 3 >> bode(h_s); >> subplot(,,) 6

8 >> step(h_s); >> subplot(,,) >> impulse(h_s); Bode Diagram Magnitude (abs) Phase (deg) Frequency (rad/sec) Step Response Amplitude Amplitude ime (sec) Impulse Response ime (sec) (d) hese filters are useful for manipulating the phase of the spectral components in a signal, without altering its magnitude. 7