DIGITAL SIGNAL PROCESSING IV

Transcription

1 DIGITAL SIGNAL PROCESSING IV MODULE CODE: EIDSV4A STUDY PROGRAM: UNIT AND UNIT VUT Vaal University of Technology

2 EIDSV4 Chapter Discrete Systems and Signals Page -. DISCRETE SYSTEMS AND SIGNALS. Sampling of analog signals Imagine a voice signal, originating from a microphone, that must be sampled for processing in a digital computer. This process is shown in Figure -. x(t) y[] A/D T sec T T 3T 4T 5T t Figure - The analog signal x(t) is sampled every T sec. and the analog values x(), x(t), x(t), x(3t),.., are converted by an analog to digital converter into the corresponding sequence of numbers y[], y[], y[], y[3],...etc. The sampled values are represented by the discrete function y[], which is a function of the discrete time variable. (For continuous functions f(t), round bracets are used, and for discrete functions f[], square bracets are used) The question now is, how much of the information (frequencies), that was originally contained in x(t), was preserved in the sampled replica y[]. Let us assume that x(t) was a voice signal in a telephone system with frequencies that vary between 3 and 3 Hz. The frequency spectrum X(f) of x(t), could be represented by Figure -. The reason for also having negative frequencies, is rooted in the fact that any signal component Acosft, contained in x(t), may be written as the sum of a positive rotating frequency and a negative rotating frequency: Acos ft A ej ft A e j ft. X(f) Figure f Remarably, given that the sampling frequency f s /T hertz, it may be shown that the frequency spectrum Y(f) of the sampled values y[], is exactly the same as the frequency spectrum X(f) of the original signal x(t), except that it will repeat every f s hertz, as shown in Figure -3.

3 EIDSV4 Chapter Discrete Systems and Signals Page - Y(f) Figure -3 -f s -3 -f s -f s f s -3 f s f s +3 f It is clear from Figure -3 that it is theoretically possible, given that f s 6 Hz., to recover the original signal x(t) from the sequence y[], with the proper low pass filter and digital to analog converter. Shannon-Nyquist sampling theorem: An analog signal containing components up to some maximum frequency, f hertz, may be completely represented by regularly-spaced samples, provided the sampling rate is at least f samples per second. (Note: If f s is the sampling frequency, ½f s is sometimes called the Nyquist frequency) A classical phenomena that can be explained with the aid of spectrum repetition, occurs in old cowboy movies where wagon wheels sometimes seem to rotate bacwards. This situation is depicted in Figure -4, where for example, a rotating wheel is sampled (filmed) at 5 frames per second while the wheel is rotating at 4, 5 and 6 revolutions per second. (Keep in mind that we have assumed that the wheel has one spoe, where as a real world wagon wheel, has many spoes, and that rotation from one spoe position to the next, will resemble one revolution.) wheel rotates at 4 revs/sec alias of 4 r/s: wheel apparently rotates bacward at r/s. 4r/s f s 5 r/s wheel rotates at 5 revs/sec alias of 5 r/s: wheel apparently stands still 5r/s f s 5 r/s wheel rotates at 6 revs/sec alias of 6 r/s: wheel apparently rotates forward at r/s. 6r/s f s 5 r/s Figure -4

4 EIDSV4 Chapter Discrete Systems and Signals Page -3 Reconstruction Consider the simple situation where a signal x(t) is converted into the numbers y[] which are immediately converted bac into the analog signal x r (t). Ideally x r (t) and x(t) should be the same. Unfortunately this can never be accomplished exactly and we must be satisfied with x r (t) only an approximate version of x(t). x r (t) can however be made to closely resemble x(t) by using a high sampling rate, precision analog to digital and digital to analog converters as well as high quality low pass filters. The reconstruction process in practice, typically uses a digital to analog converter, followed by a first order hold action (staircase), as shown in Figure -5. y[] Digital to analog converter and first order hold circuit x r (t) T T 3T 4T 5T t Y(f) a(f) sin( ft) ft X r (f) -f s f s f -f s Figure -5 f s f Although the first order hold circuit acts as a low pass filter to a certain degree, it is still necessary to implement a low pass filter at the output of the system if one wishes to obtain a good replica of the analog signal that was originally represented by the discrete numbers.. Elementary digital signals.. Stepfunction The stepfunction u(), is defined as: u[], < u[]

5 EIDSV4 Chapter Discrete Systems and Signals Page -4 u[] u[ ] u[+] u[ ] 3u[ ] 3.. Impulse function The impulse function (), is defined as: ], []

6 EIDSV4 Chapter Discrete Systems and Signals Page -5 [] [ ] Example - Setch the function (.5) {u[] u[ 3]}..5.5 Example - Setch the function (cos){u[] u[ 3]}, with in radians Example -3 Setch the function e j in the complex plane for,,, 3. : e j : e j {remember Euler s rule: re j rcos+rsin r} : e j 3: e j3 3

7 EIDSV4 Chapter Discrete Systems and Signals Page -6 Im 3 Re.3 Difference equations The relationship between the output y[] and the input x[] of a discrete system, is described by a difference equation. This is in contrast with continuous systems that are described by differential equations. An important element in discrete systems is the delay operator, represented by the symbol D in the time domain. If the input to the delay operator is x[], then the output is x[ ]. (The z transform will be discussed in Chapter 3, and the frequency domain delay operator, z -, is sometimes loosely used as the delay symbol.) x[] D (z - ) x[ ] Dx[] x[ ] z x[] x[ ] Example -4 A loan of R is obtained from a ban. The outstanding amount is augmented by an interest of % at the end of each month. a) Determine a difference equation for the amount y[] owed to the ban during month. b) Draw a bloc diagram of the system. c) Determine a solution for y[]. a) y[] y[] y[] +.y[].y[] y[] y[] +.y[].y[] y[3] y[] +.y[].y[] y[] y[ ] +.y[ ].y[ ] (the outstanding amount y[] during the present month, equals the outstanding amount y[ ] for the previous month, plus % of the outstanding amount y[ ] of the previous month)

8 EIDSV4 Chapter Discrete Systems and Signals Page -7 b). y[ ] y[]. y[ ] Delay operator. y[ ] z - y[] c) y[].y[]. [y[] initial loan] y[].y[]. [.] (.) y[3].y[]. [(.) ] (.) 3 y[4]. y[3]. [(.) 3 ] (.) 4 The solution for y[] is therefore given by: y[] (.), for y[] month.4 Linear time-invariant systems A single input, single output discrete system, may be represented as in Figure -6. x[] Discrete system Figure -6 y[] The relationship between y[] and x[] for a discrete system, is described by a difference equation. The systems that will be studied in this course will satisfy the following requirements:

9 EIDSV4 Chapter Discrete Systems and Signals Page -8. Linearity: A linear system satisfies the principle of superposition. If y [] is the response to an input x [], and y [] is the response to an input x [], then {y [] + y []} is the response to {x [] + x []}. Example -5 Determine whether the systems represented by the following difference equations are linear: a) y[] x[] +.5x[ ] b) y[].5y[ ] + x[ ] use the D operator or the z operator: y[ ] Dy[] or y[ ] z - y[] c) y[] x[]x[ 3] (Note: Apply an input x [] to the system and determine the corresponding output y []. Then apply an input x [] to the system and determine the corresponding output y []. Finally apply an input x 3 [] x [] + x [] to the system and determine the corresponding output y 3 [] for this combined input. For the system to satisfy the principle of superposition, it must be shown that y 3 [] y [] + y [].) a) x[] x []: y [] x [] +.5x [ ] x[] x []: y [] x [] +.5x [ ] So for x[] x [] + x []: y 3 [] {x [] + x []} +.5{x [ ] + x [ ]} y 3 [] {x [] +.5x [ ]} + {x [] +.5x [ ]} y [] + y []. Therefore the system is linear. b) x[] x []: y [].5y [ ]+x [ ].5z - y []+x [ ] y [] x [ ]/[.5z - ] x[] x []: y [].5y [ ]+x [ ].5z - y []+x [ ] y [] x [ ]/[.5z - ] x[] x [] + x []: y 3 [].5y 3 [ ]+x [ ]+x [ ].5z - y 3 []+x [ ]+x [ ] y 3 [][.5z - ] x [ ]+x [ ] y 3 [] x [ ]/[.5z - ]+x [ ]/[.5z - ] y 3 [] y [] + y [] The system is linear. c) x[] x []: y [] x []x [ 3] x[] x []: y [] x []x [ 3] x[] x [] + x []: y 3 [] [x [] +x []][x [ 3]+x [ 3]] x []x [ 3] + x []x [ 3] + x []x [ 3] + x []x [ 3] y 3 [] y []+ y [] The system is non-linear. Time-invariance: The properties of the system do not change with time. If y[] is the response to the input x[] then y[ ] must be the response to the input x[ ], in order for the system to be time invariant. Other system properties. Causality: A system is causal if the present output is not dependent on future inputs.. Stability: A system is stable if a limited (bounded) input results in a limited (bounded) output.

10 EIDSV4 Chapter Discrete Systems and Signals Page -9 Exercise - Write down the first ten terms of the Fibonacci series: y[], y[] and y[+] y[+] + y[]. - A signal x[] is shown in Figure P -. Setch the following signals: a) x[ ] b) x[3 ] c) x[ ]u[] d) x[ ][] e) x[ ][ ] x[] Figure P - -3 Setch the following functions: a) u[ ] b) u[+] + [] c) u[+] u[3 ] -4 Setch the following functions by calculating x[] for a range of values of.: a) x[]3e -/ b) x[]e /5 c) x[]sin(/6)u[] d) x[]e -/4 cos(/)u[] -5 Draw a bloc diagram representing the following difference equation: y[].65y[ ].934y[ ] +.5x[].x[ ] (Use the z operator (see Chapter 3): previous value z - current value.) -6 Determine whether the following difference equations represent linear systems: a) y[] x[] + x[ ] + 3x[ ] b) y[] {x[]} Solutions -) y[] y[] y[] y[3] y[4] 3 y[5] 5 y[6] 8 y[7] 3 y[8] y[9]34 y[]55 -) a) b) x[ ] x[3 ] c) d) x[ ]u[] x[ ][]

11 EIDSV4 Chapter Discrete Systems and Signals Page - e) x[ ] x[ ][ ] -3) a) b) u[ ] u[+]+[] - c) u[+] u[3 ] u[+] u[3 ] ) a) x[]3e -/ x[]... b) x[]e / x[]...

12 EIDSV4 Chapter Discrete Systems and Signals Page - c) d) x[]sin(/6)u[] x[]e -/4 cos(/)u[] x[] x[] -5) x[].5 y[] z - z - x[-].65 y[-] z - z - x[-] y[-] -6) a) y [] x [] + x [ ] + 3x [ ] and y [] x [] + x [ ] + 3x [ ] y 3 [] {x [] + x []} + {x [ ] + x [ ]} + 3{x [ ] + x [ ]} {x [] + x [ ] + 3x [ ]} + {x [] + x [ ] + 3x [ ]} y + y b) y [] {x []} and y [] {x []} y 3 [] {x [] + x []} {x []} + x []x [] + {x []} {x []} + {x []} + x []x [] y [] + y [] + x []x [] y + y

13 EIDSV4 Chapter Time Domain Analysis Page -. TIME DOMAIN ANALYSIS. Description of digital signals with impulse functions A portion of a digital signal x[], is shown in Figure -. x[] Figure - It is clear that x[] may be considered as the summation of the basic impulse functions shown below: x[][] x[] x[][ ] x[][ ] x[3][ 3] x[4][ 4] For example, the discrete function x[] u[], may also be expressed as: x[] [] + [ ] + [ ] + 3[ 3] + 4[ 4] + 5[ 5] + x[] x[] x[3] x[4] - 3 4

14 EIDSV4 Chapter Time Domain Analysis Page - Any function x[] can thus be expressed as the sum of weighted and shifted unit impulses: x[] x[][] + x[][ ] + x[][ ] + x[3][ 3] + x[4][ 4] +.. n or more compactly: x[] x[n] [ n] n n For example, x[4]. x[n] [4 n] n The only term that is not zero, x[n][4 n], occurs when n 4 and it's value is x[4]. This illustrates the important sifting property of the unit impulse function.. Impulse response Die impulse response of a digital processor is the reaction of the system when an impulse [] is applied to the input of the processor. Because [] only exists at, any output signal observed after (see Figure -), is because of the character of the system itself. For this reason the resulting response is often called the natural response of the system. The impulse response plays such an important role in the analysis of digital signal processing (DSP) systems, that it is denoted by the special symbol h[]. x[] [] y[] h[] DSP system Figure - Example - Determine the impulse response of the system below. x[] + y[] z y[] x[] + y[ ] h[] [] + h[ ] and h[] for < h[] + h[] + [] h[] h[] + System etc.

15 EIDSV4 Chapter Time Domain Analysis Page -3 Example - Determine the first four terms of h() for the system below. x[] + y[].9 z y[] x[].9y[ ] h[] [].9h[ ] h[] h[].9.9 [] h[].9(.9).8 h[3] Digital system.3 Convolution The input x[] to a system as well as impulse response h[] of the system is shown in Figure -3. x[] 3 Figure -3 Determine now the response of the system to each component of x[] individually. Consider first x[] which is an impulse of strength one at. [] Consider secondly x[] which is an impulse of strength at. [ ] h[] h[] h[ ] 4 h[] Finally consider x[] which is an impulse of strength 3 at.

16 EIDSV4 Chapter Time Domain Analysis Page -4 3[ ] 3 3h[ ] 6 3 The total response is therefore the following: y[] y[] x[]h[] + x[]h[ ] + x[]h[ ] (Note: Students should very carefully inspect why y[] could be written in this form. Also verify that y[], y[], etc, generates the correct values, for instance, y[] x[]h[] + x[]h[ ] + x[]h[ ] ) In general then the response y[], of a system at time instant, may be expressed as: y[] x[]h[] + x[]h[ ] x[ ]h[] + x[]h[] n y[] n x[n]h[ n] Equation - (a) Per symmetry, Equation - (a) could also be written as: n y[] h[n]x[ n] Equation - (b) n Equation - is valid if h[] for <, and x[] for <. Although we will always insist that h[] for < (otherwise we would really live in a terrifying world), the input x[] may exist for <. In this case if x[] for <, y[] may be expressed as: y[] n h[n]x[ n] Equation - After completion of this chapter, students are encouraged to show the validity of Equation.. Equations - and -, are very important and is nown as the convolution sum. The convolution sum is denoted by y[] x[]h[] h[]x[].

17 EIDSV4 Chapter Time Domain Analysis Page -5 In the discussion that follows, we will use Equation -(a) as basis, but we could have used the form in Equation -(b) just as well. Equation - possesses an important graphical interpretation. Assume we want to calculate y[] in the previous example. From Equation - (a) we have: y[] x[n]h[ n] x[]h[] + x[]h[] n This equation suggests that we must multiply the values of the function x[n] with the values of the function h[ n]. Drawing a graph of x[n], if x[] is given, is a simple matter because x[n] is identical to x[], as shown below, except for a change in independent variable from to n. Drawing a graph of h[ n], given h[], is a bit more tricy. A simple rule may however be used to construct for instance h[ n]. From the h[] graph, shift the vertical axis one unit to the right and mirror the graph around the vertical axis. Replace the variable by n on the horizontal axis and label the resulting graph h[ n]. x[] 3 x[n] 3 n h[] h[ n] n x[n] h[ n] 4 y[] + 5 {Add the values of the graph x[n]h[ n]} n Similarly y[] may be found by multiplying x[n] with h[ n] and adding the products. x[n] 3 h[ n] n n y[] ( )

18 EIDSV4 Chapter Time Domain Analysis Page -6 Example -3 x[] + z - y[].77 z -.5 a) Determine a difference equation for y[] and x[]. b) Calculate the first seven terms ( tot 6) of the impulse response of the system. c) Use the approximate impulse response found in b) and the method of the convolution sum, to determine the response of the system for x[]u[] u[ 3]. a) From the bloc diagram, y[] x[] +.77y[ ].5y[ ] b) To determine the impulse response, tae x[] [], so that y[] h[]. The impulse response is determined with the system initially at rest, that is to say y[ ] and y[ ]. Therefore: h[] h[] h[] h[3] h[4] h[5] +.77(.65) h[6] +.77(.449).5(.65) c) x[] h[] x[n] h[ n] n n

19 EIDSV4 Chapter Time Domain Analysis Page -7 : n x[n] h[ n] y[] : n x[n] h[ n] y[].77 : n x[n] h[ n] y[] n x[n] h[3 n] y[3].957 4: n x[n] h[4 n] y[4].875 5: n x[n] h[5 n] y[5] : n x[n] h[6 n] y[6] -.3 7: n x[n] h[7 n] y[7] : n x[n] h[8 n] y[8] -.56

20 EIDSV4 Chapter Time Domain Analysis Page -8 Exercise - Describe the signals below with the aid of weighed and shifted impulse functions. x[] {Ans: x[] [] + [ ] + [ ] x[] [+] + [].5[ ] + [ 4]} - Setch the first ten terms of the impulse responses of the digital filters described by the following difference equations. a) y[] x[] + x[ 4] + x[ 8] b) y[] y[ ] + x[] x[ 8] c) y[] y[ ].5y[ ] + x[] x[] {Ans: a) h[], h[], h[], h[3], h[4], h[5], h[6], h[7], h[8], h[9] b) h[], h[], h[], h[3], h[4], h[5], h[6], h[7], h[8], h[9] c) h[], h[], h[].5, h[3], h[4].5, h[5].5, h[6].5 etc. or as we will see in Chapter 3, h[].44(.77) cos[( )/4]} -3 Use graphical convolution to determine the output y[] for the impulse response h[] and input x[] shown below. a) x[] h[].5 b) x[] h[] Ans: a) y [ ] b) y [ ] -4 Find an expression for y[] x[]h[] if x[] u[] and h[] u[]. {Ans: y[] [ + + ]/[ ]}

21 EIDSV4 Chapter 3 Z transform Page 3-3. Z TRANSFORM 3. Definition of the z-transform The z transformation of a digital signal x[], with x[] for <, is defined as: X(z) Z{x[]} x[] + x[]z + x[]z + x[3]z X(z) x[]z Equation 3- Example 3- Determine the sum of the following sequences, a) ½ + ¼ + ⅛ and b) z + z + z 3 + z a) Let S (/) + (/4) + (/8) + (/6) +... b) Let S z + z + z 3 + z ½S (/4) + (/8) + (/6) +... S /z + /z + /z 3 + /z S ½S ½ (/z)s /z + /z 3 + /z ½S ½ S (/z)s (/z) S S[ (/z)] (/z) S (/z)/[ (/z)] Example 3- Determine the z transform of the function x[] []. X(z) + z + z +... X(z) X(z) Example 3-3 Determine the z transform of the function x[] u[]. X(z) + z + z + z X(z) + /z + /z + /z (/z)x(z) /z + /z + /z X(z) (/z)x(z) X(z) /[ (/z)] z X(z) z z Example 3-4 Determine the z transform of the function x[] a ( and a < ). X(z) a + a z + a z + a 3 z X(z) + (a/z) + (a/z) + (a/z) (a/z)x(z) (a/z) + (a/z) + (a/z) X(z) (a/z)x(z) X(z)[( (a/z)] X(z) /[ (a/z)] z X(z) z a x[] x[] x[]

22 EIDSV4 Chapter 3 Z transform Page 3- Example 3-5 Determine from first principles, the z transform of x[] X(z) + z + z + 3z 3 + 4z z X(z) z + z 3 + 3z X(z) z X(z) z + z + z 3 + z ( )X(z) z z z X(z) z z z X(z) z z X(z) z (z ) + z + z 3 + z z + z 3 + z But z + z + z 3 + z z Example 3-6 Determine the Z transform of x[] u[ ]. X(z) + z + z + z 3 + z z ( + z + z + z 3 + z ) z z (refer to Example 3-3) z Z{u[ ] z Z{u[]} Example 3-7 Show that Z{x[ ]} z Z{x[]} (refer to example 3- b) Z{x[ ]} x[ ] + x[ ]z + x[]z + x[]z 3 + x[] z x[ ] + x[ ]z - + z {x[] + x[] z + x[]z +...} z Z{x[]} as we normally assume that x[ ] x[ ] 3. Some properties of the z transform. Z{x[] + y[]} X(z) + Y(z). Z{ax[]} ax(z) 3. Z{x[ n]} z -n X(z) 4. Z{x[ + n]} z n X(z) z n x[] z n- x[]... zx[n ] 5. Z{x[]y[]} X(z)Y(z) 6. x[] X(z) z Lim z 7. x[] X(z) z Lim z (Initial value theorem) (Final value theorem) Justification:. Z{x[] + y[]} {x[] + y[]}z x[]z y[]z X(z) + Y(z). Z{ax[]} ax[]z a x[]z ax(z) x[] x[]

23 EIDSV4 Chapter 3 Z transform Page Z{x[ n]} x[ n] + x[ n]z + x[ n]z n x[ ]z + x[]z n + x[]z n + x[]z n +... Now, we assume that x[] for <, so that x[ n] x[ n] x[ n]... x[ ] Z{x[ n]} + z + z z n+ + x[]z n + x[]z n + x[]z n +... x[]z n + x[]z n + x[]z n +... z n {x[] + x[]z + x[]z +...} z n X(z) 4. Z{x[ + n]} x[n] + x[n + ]z + x[n + ]z +... z n {x[n]z n + x[n + ]z (n+) + x[n + ]z (n+) +...} z n {x[] + x[]z x[n ]z (n ) + x[n]z n + x[n + ]z (n+) +...} z n x[] z n x[]... zx[n ] z n X(z) z n x[] z n x[]... zx[n ] 5. x[]y[] x[]y[][]+{x[]y[]+x[]y[]}[ ]+{x[]y[]+x[]y[]+x[]y[]}[ ] + {x[]y[3] + x[]y[] + x[]y[] + x[3]y[]}[ 3] +... Z{x[]y[]} x[]y[] +{x[]y[] + x[]y[]}z +{x[]y[] + x[]y[] + x[]y[]}z + {x[]y[3] + x[]y[] + x[]y[] + x[3]y[]}z Now: X(z)Y(z) {x[] + x[]z + x[]z + x[3]z }{y[] + y[]z + y[]z + y[3]z } x[]y[] + x[]y[]z + x[]y[]z + x[]y[3]z x[]y[]z + x[]y[]z + x[]y[]z x[]y[]z + x[]y[]z x[3]y[]z x[]y[] + {x[]y[] + x[]y[]}z + {x[]y[] + x[]y[] + x[]y[]}z + {x[]y[3] + x[]y[] + x[]y[] + x[3]y[]}z Z{x[]y[]} X(z)Y(z) 6. X(z) z Lim Lim x[] x[] x[] x[3]... z z z z 3 x[] z 7. Let Y(z) [(z )/z]x(z) so that X(z) z Lim Y(z) z z Lim Y(z) [ z ]X(z) X(z) z X(z) x[]z x[ ]z Y(z) z Lim z Lim x[]z x[ ]z x[] x[ ] {x[] + x[] + x[] +...} {x[ ] + x[] + x[] + x[] +...} x[] x[ ] + (all terms cancel except for the final term x[], of the x[] sequence and x[ ]. Lim z So, assuming that x[ ], we may conclude: x[] X(z). z z We must add however, that x[] will only reach a stable final constant value if X(z) has either a single pole at z and all other poles reside within the unit circle or if all poles are contained within the unit circle (in which case the final value is zero).

24 EIDSV4 Chapter 3 Z transform Page 3-4 Example 3-8 Determine the z transform of x[] 3u[] + (.5) u[]. X(z) Z{3 + (.5) } Z{3} + Z{(.5) } 3z/(z ) + z/(z.5) [3z(z.5) + z(z )]/(z )(z.5) z(4z.5)/(z )(z.5) Example 3-9 Calculate the z transform of x[] u[ 3]. X(z) Z{u[ 3]} z -3 Z{u[]} z -3 [z/(z )] z - /(z ). Example 3- The input to a system with impulse response h[] (.5) u[], is x[] u[]. Determine the z transform Y(z), of the output y[] x[]*h[]. x[] u[] X(z) z/(z ) and h[] (.5) u[] H(z) z/(z.5) Y(z) X(z)H(z) [z/(z )][z/(z.5)] z /(z.5z +.5). Example 3- Determine the z transform of the function x[] sin()u[]. e jx e jx From Euler s rule we have sinx, j e X(z) Z{sin} jθ x[] ejθ Z j j j j z j jθ jθ Ze e Ze jθ e jθ z z see example 3-4 z e jθ z e jθ jθ jθ (z e ) (z e ) (z e jθ )(z e jθ ) z j e jθ e jθ z j (z e jθ )(z e jθ ) jθ jθ e e (z e jθ )(z e jθ ) z sinθ (z e jθ )(z e jθ ) z sinθ (a very important form) (z e jθ )(z e jθ ) z(sinθ) z(sinθ z(sinθ) z ze jθ ze jθ z z [(e jθ + e jθ )/] z z(cosθ) z sinθ z sinθ z sin θ So, X(z) z or X(z) z(cosθ) (z e jθ )(z e jθ ) (z θ)(z θ)

25 EIDSV4 Chapter 3 Z transform Page 3-5 Example 3- Show that Z{ab cos( + )} Z{ab cos( + )} j j - j j (az (a z. z b z b ab e e cos( + ) ab j( ) j( ) e jx e jx (from Euler s rule, cosx j( ) j( ) ab e e ae jbe j aejbe j Z ae be ae be ae j Z be j ae j Z j be j z j z (az (a z ae ae z be j z bej z b z b (using the normal notation for complex numbers. re j r) ) Summary of z transforms: x[] X(z) [] u[] z (z ) a u[] z (z a) u[] (z z) a u[] az (z a) sinu[] z(sin) (z e j )(z e j ) cosu[] ab cos( + ) x[ ] x[ ] x[ + ] x[ + ] z -z(cos) (z e j )(z e j ) (a z (az z b z b z X(z) z X(z) z X(z) z x[] z X(z) z x[] z x[]

26 EIDSV4 Chapter 3 Z transform Page Inverse z transform 3.3. Long division Example 3-3 Determine x[] if X(z) z/(z.5) +.5z - +.5z - +.5z z.5 z z z -.5z -.5z - -.5z -.5z -.5z z -3 x[], x[].5, x[].5, x[3].5,...x[] (.5) Example 3-4 Determine the inverse z transform of X(z) (z )/(z.9z +.84) z - +.9z z z.9z +.84 z z z z -.9.7z z -.87z -.756z -.87z -.653z z z -.738z -3 x[], x[], x[].9, x[3].87, Method of partial fractions. This method demands that X(z) should be a proper fraction, that is, the order of the numerator must be smaller than the order of the denominator.. Preferably X(z)/z must be expanded into partial fractions. Example 3-5 (Case I - different real factors) Determine the inverse z transform of X(z) /(z )(z.5) X(z) /(z )(z.5) X(z)/z /z(z )(z.5) A/z +B/(z ) + C/(z.5) A /z(z )(z.5) z /( )(.5) B /z(z )(z.5) z /(.5) C /z(z )(z.5) z.5 /.5(.5 ) 4 X(z)/z /z + /(z ) 4/(z.5) X(z) +z/(z ) 4z/(z.5) x[] [] + u[] 4(.5) u[]

27 Example 3-6 (case II - complex factors) Determine y[] if x[] u[] for the difference equation: y[] x[] + y[ ].5y[ ] and the system initially at rest. Y(z) X(z) + z - Y(z).5z - Y(z) z Y(z) z X(z) + zy(z).5y(z) (z z +.5)Y(z) z X(z) EIDSV4 Chapter 3 Z transform Page 3-7 This is a simple method to find the complex factors of z z +.5. (If the coefficient of z is not one, then that coefficient should first be taen out as a common factor from the quadratic expression.) Set z z +.5 z [ ( 4.5)]/ [ ( )]/ z ( j)/.5 j r Therefore z z and z z Thus z z +.5 (z )(z ) (z )(z )y(z) z X(z) For x[] u[], X(z) z/(z ) (z )(z )y(z) z [z/(z )] Y(z) z 3 /(z )(z )(z ) Y(z)/z z /[(z )(z )(z )] Y(z)/z A/(z ) + B/(z ) + C/(z ) A z /(z )(z z +.5) z /( +.5) {utilizing the pre-factored form Y(z)/z z /(z )(z z +.5)} B z /(z )(z )(z ) z ( ) /( )( ).5.57/( )(.57) C z /(z )(z )(z ) z ( ) /( )( ).5.57/( )(.57) B * (we will always find that C B * and therefore we will never calculate C in future) Y(z)/z /(z )+( )/(z )+( )/(z ) Y(z) z/(z )+( )z/(z )+( )z/(z ) y[] + ( )( ) + ( )( ) +( )[(.77).7854] + ( )[(.77).7854] + (.77)(.77) [.356(.7854) +.356(.7854)] + (.77)(.77) [( ) + ( )] +.77(.77) cos( ) {+ cos} y[] +.44(.77) cos( ) Two complex roots will always result in a time function of the form: ab cos(+), and in future, after finding the partial fractions, we will directly use the following rule: (az (a z z b z b ab cos( )

28 EIDSV4 Chapter 3 Z transform Page 3-8 Given that the transfer function of our system is Y(z)/X(z) z /(z z +.5), the step response may be drawn with the aid of MATLAB using the following statements: a[ ] b[ -.5] xdstep(a,b,) stem(x) Example 3-7 (case III - repeating roots) Consider the system shown below and determine y[] if x[] u[]. x[] + y[] z z - y[] x[] + y[ ].5y[ ] Y(z) X(z) + z Y(z).5 z Y(z) z Y(z) z X(z) + zy(z).5y(z) (z z +.5)Y(z) z X(z) (z.5)(z.5)y(z) z 3 /(z ) Z{u()} z/(z ) Y(z)/z z /(z )(z.5) Y(z)/z A/(z ) + B/(z.5) + C/(z.5) A z /(z )(z.5) z z /(z.5) z /(.5) 4 The highest order repeating term is handled in the usual way: B z /(z )(z.5) z.5 z /(z ) z.5 (.5) /(.5 ).5/(.5).5 To determine the next lower order fraction value, the previous residue ((z.5) removed from Y(z)/z) must first be differentiated C d/dz[z /(z )(z.5) ] z.5 d/dz{z /(z )} z.5 {[z(z ) z ]/(z ) } z.5 [(z z)/(z ) ] z.5 (.5.5)/(.5 ) (.75)/(.5) 3 Y(z) 4z/(z ).5z/(z.5) 3z/(z.5) y[] 4 (.5) 3(.5)

29 Remember: d dx df dg (f g) g f and dx dx EIDSV4 Chapter 3 Z transform Page 3-9 d dx Also for R repeating roots z z r : if X(z).. + d then A n (z z r ) R X(z) f g (z df dx A R z ) r g f g + dg dx (z A R z ) r n dz n n,,... R n! zz r (A R d (z z r ) X(z), A (z z R r ) X(z) zz r dz zz r (z z ) r x[] x[] x[] x[3] x[4] A R d R, A (z z r ) X(z)! dz, etc) zz r Exercise 3- Develop (long division) the following z transforms in power series of z -, and write down the first five sample values of x[]. a) X(z) /(z.5) b) X(z) z/(z +.) c) X(z) (z + )/(z ) {Ans c): + z - + z - + z -3 + z z z + z z - z - z - z - z - z - z -3 z Two digital signals are given by: x [] [] [ ] + [ 3] and x [] [ ] + [ ] [ 3] Determine X (z) and X (z). Convolute the two signals to form a third signal x 3 [] and show that Z{x 3 []} is equal to X (z)x (z). 3-3 Determine the 'th term of the Fibonacci series: y[+] y[+] + y[], y[], y[] {,,,, 3, 5, 8, 3,,...} {Ans: y[+] y[+] y[] [z Y(z) z y() zy()] [zy(z) zy()] Y(z) (z z )Y(z) z + z z (z z )Y(z) z Y(z)/z /(z z ) /[z ( + 5)/][z ( 5)/] A/[z ( + 5)/] + B/[z ( 5)/] z z z [ ± ( + 4)]/ {z [( + 5)/]} and {z [( 5)/]} A /[z ( 5)/] z(+5)/ /[( + 5) ( 5)/] /5 B /[z ( + 5)/] z( 5)/ /[( 5) ( + 5)/] /5 Y(z) (/5)z/[z ( + 5)/] (/5)z/[z ( 5)/] y[] (/5)[( + 5)/] (/5)[( 5)/] y[] (/5){[( + 5)/] [( 5)/] y[] (/5){ ( ) }, where is the golden number.683 and the reciprocal of, /.683} {chec: y[], y[], y[], y[3], y[4] 3, y[5] 5, y[6] 8...} 3-4 Use z-transform methods to find y[] from the following difference equation: y[].6y[ ].5y[ ] + x[] if x[] []

30 EIDSV4 Chapter 4 Frequency Domain Analysis Page 4-4. FREQUENCY DOMAIN ANALYSIS 4. Transient response and steady state response The complete response of a system consists of the transient behaviour and the steady state behaviour. The response of a system with no driving force (x[] ) but with initial conditions, is called the transient response. The forced or steady state response depends on the specific input signal. The transient response is intimately connected to the impulse response h[], specifically the location of the roots of the denominator (also called the poles) of H(z), as illustrated in Figure 4-. Inside unit circle b < : y[] Ab + b Real roots: z b Outside unit circle b > : x[] Ab + b eg. (-½) eg. (½) eg. (-) eg. () Real roots or, z or z + y[] A( ) + or y[] A() + Roots residing on the unit circle: z Complex roots (z )(z -) y[] Acos +. (-) () Inside unit circle: b < y[] Ab cos + Complex roots: (z b)(z b ) Outside unit circle: b > y[] Ab cos + b Decaying sinusoid b Growing sinusoid Figure 4-

31 EIDSV4 Chapter 4 Frequency Domain Analysis Page 4-4. System transfer function and frequency response If the impulse response of a discrete system is given by h[] and the z transform of h[] is H(z), then the frequency response H() of the system is given by: H() H(z) z e j T jt h[] e Equation 4- To derive this result, consider the digital system depicted in Figure 4-. x(t) x[] y[] x[]*h[] h[] H(z) Figure 4- T X(z) Y(z) X(z)H(z) We will now apply the signal x(t) sint with unit amplitude, which will, after sampling at frequency s /T, become the digital signal x[] sint u[]. We may find the output y[], using the z transform techniques from Chapter 3. With x[] sint u[], then X(z) zsint/(z e jt )(z e jt ) zsint Y(z) H(z) X(z) Y(z) H(z) jt jt ( z e )( z e ) Y(z)/z H(z)sinT/(z e jt )(z e jt ) Y(z)/z A/(z e jt ) + A * /(z e jt ) + terms with poles of H(z) (Note: Although we do not now the exact expression for H(z), we do now that for a stable system H(z), the poles of H(z) must lie within the unit circle (refer to the illustration in Figure 4-), and the time domain representation of the terms corresponding to the poles of H(z) (the natural response), will vanish. The z transform, Y ss (z), of the steady state output y ss [], will therefore only involve the poles of X(z).) Y ss (z)/z A/(z e jt ) + A * /(z e jt ) A [H(z)sinT]/[(z e jt )] z e j T [H(e jt )sint]/[(e jt e jt )] [H(e jt )sint]/{j[(e jt e jt )/j] [H(e jt )sint]/[jsint] {from Euler s rule: sint (e jt e jt )/j} H(e jt )/j A H(e jt ) / and A H(e jt ) j H(e jt ) / Y ss (z) Az/(z T) + A * z/(z T) { e jt T } Using now (a)/(z b) + (a )/(z b ) ab cos( + ), with a H(e jt ) /, H(e jt ) /, b and T: y ss [] ( H(e jt ) /)() cos[t + H(e jt ) /] y ss [] H(e jt ) sin[t + H(e jt )] {cos( /) sin} The frequency response of a discrete system, may therefore be determined from H(z) by replacing z with e jt. The frequency response of a discrete system, often just denoted by H(), is therefore: H(e jt ) H(z) e j z T jt h[]e x[] sint y[] H sin(t+h) h[] H() H()

32 EIDSV4 Chapter 4 Frequency Domain Analysis Page 4-3 Example 4- A system function is given by H(z) z(z )/(z.44z +.64) a) Determine a difference equation for the system b) Draw a bloc diagram of the system c) Determine an expression for h[] d) Draw a graph of the frequency response of the system for T to e) Calculate the steady state response y ss [] of the system, if the input to the system is x[] 5sin(3+) a) Y(z)/X(z) z(z )/(z.44z +.64) (z.44z +.64)Y(z) z(z )X(z) (z z)x(z) z Y(z).44zY(z) +.64Y(z) z X(z) zx(z) Y(z).44z Y(z) +.64z Y(z) X(z) z X(z) y[].44y[ ] +.64y[ ] x[] x[ ] y[] x[] x[ ] +.44y[ ].64y[ ] b) x[] + y[] z z.44 z c) H(z) z(z )/(z.44z +.64) H(z)/z (z )/(z.44z +.64) H(z)/z (z )/[z (.8/.5)][z (.8/.5) ] A/[z (.8/.5)] + A * /[z.8/.5) ] A (z )/(z.8.5) z.8.5 (.8.5 )/( ) / z.44z +.64 z [.44 (.44) 4.64]/ [ ]/ [.44 j.7673]/.7 j /.5 H(z) ( )z/(z.8.5) + ( )z/(z.8.5) h[].633(.8) cos( ).66(.8) cos( ) d) H(z) z(z )/(z.44z+.64) H() H(e jt ) e jt (e jt )/[(e jt ).44e jt +.64)] T(T )/[(T).44T +.64] T(T )/(T.44T +.64) H() () )/( )

33 EIDSV4 Chapter 4 Frequency Domain Analysis Page 4-4 H(/4) (/4)((/4) )/[(/).44(/4) +.64] Table 4- summarises the values for H(T), for T, /4, /, 3/4,, 5/4, 6/4, 7/4 and. Figure 4-3 is a visualization of z s movement around the circle. T z e jt H(e jt ) /4 / / / /4 3/ /4 5/ / 6/ /4 7/ T Table 4- Figure 4-3 z e jt H() is thus obtained from H(z), with z replaced by values on the unit circle that is z e jt T. The frequency response therefore repeats when T reaches that is /T which is the sampling frequency. Figure 4-4 is a rough graph of the tabulated values of H(), as well as a MATLAB plot (a[ - ];b[ ];[H,w]freqz(a,b,5,'whole'); plot(w/(*pi),abs(h)),grid). 3 H() T Figure 4-4 e) In general, if a structure has a system function H(z) and the input to the system is x[] Asin(a+), then the steady state output is given by y ss [] H(e ja ) Asin[a+ +H(e ja )] (or similarly, if x[] Acos(a+) then y ss [] H(e ja ) Acos[a++H(e ja )]) So with x[] 5sin(3+) and H(e j3 ) (e j3 )/(e j6.44e j3 +.64) it follows that y ss [] sin( ) 3.94sin(3.54)

34 EIDSV4 Chapter 4 Frequency Domain Analysis Page The non-uniqueness of digital signals Although the calculation of the discrete frequency response H() from H(z) with z replaced by e jt, may be vaguely reminiscent of the calculation of the continuous frequency response H() from H(s) with s replaced by j, it is clear from the discussion in section 4-3 and Example 4-, that the frequency response of continuous systems and discrete systems, differ dramatically. One reason for this difference is rooted in the difference between the continuous-time excitation sint and the discrete-time excitation sint. The continuous-time system is sensitive to the unique shape of sint at each frequency in radians per second while the discrete-time system is sensitive to the digital frequency T in radians per sample. What is important to understand is that two different sinusoids may still produce the same sequence of samples, if the sampling rate is changed. A discrete system will then not be able to distinguish between these identical samples, caused by two different continuous-time signals. And also, only when one has nowledge of the sampling rate or sampling frequency, is it possible to interpret the frequency response of a digital system, and this interpretation will only be valid up to the Nyquist frequency (half the sampling frequency), because after this, the frequency profile will either be a mirror image or a recurring image of the principal frequency image between and the Nyquist frequency. We will further explore some of these concepts in Chapter 5. Exercise 4-. Determine the frequency response of the system H(z) z/(z.5)(z.8), for T, /4, /, 3/4 and. 4-. Calculate H() if h[] {, 3, 5, 7, }

35 EIDSV4 Chapter 5 Discrete Fourier Transform Page 5-5. DISCRETE FOURIER TRANSFORM 5. Frequency spectrum of non periodic digital signals The frequency spectrum X() of a sequence of numbers x[], obtained when a signal x(t) was sampled every T second (sampling frequency s /T), is given by: X() jt x[]e where T / s Equation 5- The normalised form of Equation 5- results if we assume that T or that X() is essentially a function of T, the sampling angle (radians/sample): X() j x[]e Equation 5-(a) Although X() is indicative of the frequency content of a non repetitive signal x[], it is perhaps still useful, in light of the striing similarity between Equation 5- and Equation 4-, to understand X() as the frequency response" of a system with impulse response x[]. The only concession we must mae, if we consider x[] to be the impulse response of a system, is that x[] may have values for <, according to Equation 5-. Even though this can never be true in the real world, we will encounter the idea of a non-causal impulse response in Chapter 6, and we will show how to deal with this. Just a few observations, regarding Equation 5-. As is always true, the frequency image of an event that is sampled, is repetitive with period /T. From Equation 5-, X(+/T) j( ΤT x() e jτ j x() e e jτ x() e X() (e j, for all integers ) Same input to the system, same response Input T Input cos[(/t)t] Input cos[(/t)t] Input cos[(3/t)t] Input cos[(/t)t] Input cos[(5/t)t] In general, X() X(+n/T), for any integer n. The periodic nature of the amplitude spectrum in particular, serves as confirmation of our initial illustration of this phenomenon, in Figure -3. We should also note that, and this will be illustrated in Example 5-, that X() is a continuous function of. X() is continuous because x[] is assumed to be non-periodic and X() is periodic because x[] is a sampled signal. The signal x[] can again be recovered from X() with the following operation: / x[] s s s / T jt X( ) e dω where s /T Equation 5- Again, a normalised form of Equation 5- results, if we assume that T (or s ): j x[] X( ) e dω Equation 5-(a) π

36 EIDSV4 Chapter 5 Discrete Fourier Transform Page 5- Equation 5- is called the discrete time Fourier transform (DTFT) while Equation 5- is called the inverse discrete time Fourier transform (sometimes abbreviated as IDTFT). A proof that X() in Equation 5-(a) and x[] in Equation 5-(a), form a transform pair, is given in Appendix 5-A at the end of Chapter 5. This transform will be important for us when we design finite impulse response filters, discussed in Chapter 6. Example 5- Calculate the frequency spectrum X() of the signal x[] in Figure 5-. x[] Figure 5-.5 T.5 X() x[] ej T x[]e j, T and x[] for >.5e j + +.5e -j + cos (A representation of X() is given in Figure 5-.) X() cos + Figure 5- Example 5- Determine the frequency spectrum X() of the signal x[] shown in Figure 5-3. Use the inversion formula to recover x[] from X(). x[].5 T.5 Figure 5-3 jt j X() x[] e x[] e.5e j +.5e -j cos Now: x[] s / j T X( ) e d s s / j cos e d, T and s /T j j e e j e d j( ) j( ) e e dω 4 j j e e 4 j(+ ) j( ) x[] j j j j e e e e 4 j(+ ) j( ) j(+ ) j( ) j j j( j e e e e 4 + j j sin[ ) ] sin[ sin[ )] sin[ (+ ) ( ) x[] sin sin, x[] sin sin.5, x[ ] sin sin.5 sin[ ) sin[ - and x[] for all (int) > (Remember Lim sinx (+ ) (- ) x x )

37 j X( )e EIDSV4 Chapter 5 Discrete Fourier Transform Page 5-3 Example 5-3 Determine the sequence x[], with T, that correspond to X(). x[] s / jt X( ) e dω s s / j e d T s j e j j j j e e e e sin j j j when () when (confirming that the frequencies contained in an impulse, stretch from to infinity) Example 5-4 Determine the sequence x[], with T, that corresponds to X() shown in Figure 5-4. X() Figure 5-4 / /4 /4 / x[] j X( )e d j e d + j e d j e j j/ e j j + e j j/ j/4 e e j j + j/4 e j j/4 e j j/ e j + j/ e j j/ j/ e e j j/4 e j j/4 j/4 e e j (/)(sin/) (/)sin(/4) ½[(sin/)/(/)] ¼[(sin/4)/(/4)] Writing x[] in the form x[] ½[(sin/)/(/)] ¼[(sin/4)/(/4)], will facilitate the calculation of x() ½ ¼ ¼, using sin x x Lim Example 5-5 Determine the sequence x[] (T ), that corresponds to X() shown in Figure 5-5. Use the result from integral calculus: xe ax dx eax a (x a ). X() Figure 5-5 X() π π π π x[] d j j j e d e d e d + j e d + + e j e j ( ) + e j ( + e j a ea using j j j j j j e d ( ) a a e j e j/ e e j e [ ( ) ( ) ] + j e ( ) ( ) + e j / e j j j j j j j j j j j j j e j e j/ + e j e j + e j j j j j j j e j + + e j / e j j j j j e j/ + e j + e j + e j / e j / e j/ e j e j + + j j j sin( /) + sin + cos sin + (cos ) cos /) The form x[] ½[sin(/)/(/)] ½ [( cos)/ ] will facilitate the calculation of x[] ½ ½ Lim cosx (½).4747, using the result from the theory of limits: x x x

38 EIDSV4 Chapter 5 Discrete Fourier Transform Page Frequency description of periodic sequences A periodic digital signal x[], obtained by sampling a periodic continuous signal x(t) every T seconds, with a total of N samples per period, is shown in Figure 5-6. x[] T Period NT N- N Figure 5-6 In a similar way that we used the exponential Fourier series to resolve a periodic continuous signal into distinct spectral frequencies, so too can x[] in Figure 5-6 be decomposed into its constituent sinusoidal frequencies (although digital in nature), starting with a fundamental frequency (e j(/nt)t ) of /NT (rad/sec), then a second harmonic (e j(/nt)t ) with frequency (/NT) and so on. For a continuous periodic signal x(t), there may be an infinite number of frequency components, but this is not true when x[] is the sampled representation of x(t), as shown in Figure 5-7. Continuous signal Digital signal Figure 5-7 N First of all, because x[] is sampled at a frequency of /T, the frequency image of this signal must repeat with a frequency period of /T. This means that there can only be N frequency components, as N(/NT) will bring us bac to the sampling frequency /T. Secondly, only half of these N components will have any real meaning as the other half will be their conjugate mirror image that we need to resolve the complex frequencies into real frequencies. This is not really a surprise, because we should have, in any case, obeyed Shannon s law by removing and filtering all frequencies from x(t), above half the sampling frequency, before sampling commenced. So the highest frequency contained in x[] should only be half the sampling frequency. Figure 5-8 could conceivably, with N assumed to be an uneven integer, depict the complex amplitudes X[n] of the frequencies contained in a given x[]. The integer n (,,... N) indicates the number of the harmonic and at the same time also the corresponding frequency (/NT, (/NT),..., /T). X[n] ( N ) NT N N N N N Period /T The amplitude spectrum X[n], of an N sample periodic digital signal, is now defined by, N - j(/n)n X[n] x[]e NT NT n ( N ) NT 3 NT ( N 3) NT ( N ) NT ( N ) NT, n,,..., N- Equation 5-3 NT T n n Figure 5-8

39 EIDSV4 Chapter 5 Discrete Fourier Transform Page 5-5 Referring to Figure 5-8, X[] is a special value, always real and indicative of the average value of x[]. Equation 5-3 clearly implies that we do not view the complex amplitudes as a symmetrical spread around zero. This doesn t really matter, due to the periodic spacing of X[n]. Let us show for example, that X[N ] X[ ]. Ν Ν X[N ] x[] e j ( x[] ej( ) ej()( ) x[] ej( )( ) X[ ] (e j and assuming Eqn. 5-3 holds true for negative n) Before we give the expression for reconstructing x[] from its constituent complex sinusoids e jn(/nt)t and associated complex amplitudes X[n], just a few remars. Firstly, consider the arbitrary periodic signal x(t) in Figure 5-9 (a), with period T. Let us assume that x (t) in Figure 5-9 (b), is the fundamental harmonic contained in x(t). x(t) Figure 5-9 Consider now x[] in Figure 5-9 (c), the same signal x(t), sampled N times per period with sampling period T (be careful, different meanings of T in the continuous and digital domain) and with total period of NT. Assume, just for the moment, that we may obtain an expression for the digital fundamental harmonic x [], in Figure 5-9 (d), by sampling x (t) in Figure 5-9(b). Therefore, we must replace T and t in x (t), with NT and T respectively, to obtain x []A sin(/n). The digital frequency /N (radians per sample) and the time index (samples), now play the same role as the continuous frequency /T (radians per second) and time index t (seconds) respectively. So, the discrete sinusoid sin(/n) is just a sequence of numbers, but in the bacground it is associated with a frequency /NT. Secondly, just to get real again, let us mae sure that if we add, for instance, the sinusoid at n and its phantom at n N, that they will add up to a real sinusoid, representing the spectral component with fundamental frequency /NT. Let us call this component, embedded in x[], the signal x []. X[n] T (a) X[]e j(/n) x [] X[]e j(/n) + X[N ]e j(/n)(n ) X[]e j(/n) + X[N ] e j e j(/n) X[]e j(/n) + X[ ]e j(/n) (e j and X[N ]X[ ] as was shown previously) Now X[] N Similarly, X[ ] t + x (t) Ν j( Ν x[] e cos(ν jx[]sin(ν} Ν Ν cos(ν (b) x (t) A sin(/t)t X[N ]e j(n )(/N) n t {x[] (n in Eq. 5-3) x[] j x[]sin( Ν (using re j rcos jrsin) Ν a jb (defining a x[] cos(ν and b Ν Ν j( Ν Ν x[] e x[] cos(ν +j Ν Ν Ν x[]sin( ) x[]sin( a + jb x [] (a jb )e j(/n) + (a + jb )e j(/n) and from Euler s rule, x [] (a jb )[cos(/n) + jsin(/n)] + (a + jb )[cos(/n) - jsin(/n)] x[] T NT (c) X[] cos{(/n)+x[]} x [] (d) x [] A sin(/nt)t A sin(/n)

40 EIDSV4 Chapter 5 Discrete Fourier Transform Page 5-6 x [] (a + a jb + jb )cos(/n) + [ja ja (j )b (j )b ]sin(/n) a cos(/n) + b sin(/n) Using the identity acos+bsin(a +b )cos[ tan - (b/a)](a +b )cos[+tan - (-b/a)], then x [] (a +b )cos[(/n)+tan - (-b /a )] X[] cos[(/n) + X[]] This in general, establishes the relationship between any complex sinusoid X n e j(/n)n and its associated real sinusoid X[n] cos[n(/n)+x[n]]. We will see in Example 5-7 that the amplitude X[n] is actually wrong, but we will discuss the necessary corrections. In conclusion then, the expression for reconstructing x[] again from X[n] and basic complex sinusoids e jn(/n), is given by, x[] N j( /N)n X[n]e,,,..., N- Equation 5-4 N n The relation given by Equation 5-3 is called the discrete Fourier transform (DFT) and the relation in Equation 5-4 is called the inverse discrete Fourier transform (sometimes abbreviated as IDFT). A bouquet of numerical algorithms exist today, with its roots primarily in the wor done by the brilliant mathematician Karl Gauss, and collectively nown as the fast Fourier transform or FFT, that greatly enhances the computational speed of the DFT and IDFT. Proof that X[n] in Equation 5-3 and x[] in Equation 5-4, form a transform pair, is given in Appendix 5-B at the end of Chapter 5 Example 5-6 Given the N 4 samples of the sequence x[] {,,, },,,, 3, determine the discrete Fourier transform X[n], and recalculate x[] again from the inverse of X[n]. N- - j(/n)n X[n] x[]e [e + e -j(/4)n + e -j(/4)n + ] X[] + + 3, X[] + e -j(/) + e -j() j, X[] + e -j() + e -j() and X[3] + e -j(3/) + e -j(3) j By means of Matlab: fft([ ]) ans 3...i..+.i The inverse may now be found with N- j(/n)n x[] X(n)e (/4)[3e +.57e j(/4) +e j(/4) +.57e j(/4)3 ] N n x[] (/4)[ ] (/4)4 x[] (/4)[3 +.57/ /] (/4)4 x[] (/4)[ ] (/4)4 x[3] (/4)[ / /] (/4) By means of Matlab: ifft([3 i i]) ans..+.i..i Example 5-7 A signal x(t) cost, with a fundamental period of sec, is sampled every T.5sec, starting at sec. A graph of the resulting sequence x[] cost cos.5, is shown in Figure 5-. The values of the N 8 samples are x[]., x[].777, x[]., x[3] -.777, x[4] -., x[5] -.777, x[6]. and x[7].777. a) Use the digital Fourier transform to calculate the values for X[] and X[].

41 EIDSV4 Chapter 5 Discrete Fourier Transform Page 5-7 b) Determine an expression for the fundamental harmonic (n ), contained in x(). c) Use the inverse digital Fourier transform to reconstruct x[] from X[n]. We now beforehand that the average value of x(t) is zero and with only the fundamental harmonic ( rad/sec), we expect spectral values for X[] and its counterpart X[7] only. x[] Figure 5- Ν a) X[n] j(n x[] e Period NT 8.5 sec 7 j(n x[] e 7 j( n x[] e X[] x[] + x[] + x[] + x[3] + x[4] + x[5] + x[6] + x[7] X[].e j(/4) +.777e j(/4) +.e j(/4).777e j(/4) 3.e j(/4) 4.777e j(/4) 5 +.e j(/4) e j(/4) /4 + / /4 -. 4/ /4 +. 6/ / j It is quite interesting to see, refer to Figure 5-, how X[] is computed as a correlation process between the testing vector e j(/4) producing X[] as the vector resultant. 4 Rotating vector e j(/4) x[] and the values of x[], Correlation between x[]cos/4 and e j(/4) 3. Figure 5- b) We found earlier that X[] a jb where a is associated with the amplitude of the cosine term and b is associated with the amplitude of the sine term. The fact that X[] is real, is then correct because we now that the fundamental frequency component is only a cosine term. However, we also now that the amplitude of the cosine term in x[] is and not 4. This discrepancy occurs because of the way the factor (/N) is used for the definition of X[n] and x(] in Equations 5-3 and 5-4. The correction factors are however very simple: ) The average value of x[] is given by: average{x[]} (/N)X[] ) The amplitude c n, of the combined real sinusoid c n cos[(/n)n + X[n]], associated with all X[n], n < N/, is given by: c n (/N) X[n] x[3]e 3j(/4) x[7]e 7j(/4) x[]e j(/4) x[4]e 4j(/4) x[]e j(/4) x[5]e 5j(/4) X[]

42 EIDSV4 Chapter 5 Discrete Fourier Transform Page 5-8 3) The amplitude a n of only the cosine term associated with all X[n], n < N/, is, given by: a n (/N)e{X[n]} 4) The amplitude b n of only the sine term associated with all X[n], n < N/, is given by: b n (/N)m{X[n]} 5) If N is even, the Nyquist component is (/N) X[N/] cos[ + X[N/]] So for the current problem, a (/8)e{4 + j}, b (/8)m{4 + j)} and c (/8) X[] (same as a for this problem because of only the cosine term). The fundamental frequency signal x [], contained in x[] is therefore given by, x [] cos[(/nt)t] cos[(/8))] cos[(/4)]. Again, all recollection of time is lost from x [] which is now just a sequence of numbers. Although we now the frequency of the fundamental component must be /NT, perhaps it is easier to visualize x [] cos[(/4)] cos(t) so that from T /4, we can conclude that (/4)/ rad/sec Hz. c) We already calculated X[] and X[], but we still need the values for X[] to X[7], in order to reconstruct x[], with the aid of Equation 5-4. We can anticipate that X[] X[3] X[4] X[5] X[6], but let us just chec X[] and X[6] X[].e j(/4) +.777e j(/4) +.e j(/4).777e j(/4) 3.e j(/4) 4.777e j(/4) 5 +.e j(/4) e j(/4) /4 + 4/ /4 -. 8/4.777 /4 +. / /4 (as expected) X[6].e j(/4) e j(/4) 6 +.e j(/4) 6.777e j(/4) e j(/4) e j(/4) e j(/4) e j(/4) /4 + / /4. 4/ / / /4 (as expected) X[7].e j(/4) e j(/4) 7 +.e j(/4) 7.777e j(/4) 7 3.e j(/4) e j(/4) e j(/4) e j(/4) /4 + 4/4.777 /4. 8/ /4 +. 4/ /4 4 (as expected) We are now ready to calculate x[] with the aid of Equation 5-4 and compare with the values of x[] that we already now. x[] N Ν n j(n X[n] e 7 8 n j(n X[n] e 7 j( n X[n] e 8 n x[] (/8)[.e j(/4) + 4.e j(/4) +.e j(/4) +.e j(/4) 3 +.e j(/4) 4 +.e j(/4) 5 +.e j(/4) e j(/4)7 (/8)[ ] (/8)[8] x[] (/8)[.e j(/4) + 4.e j(/4) +.e j(/4) +.e j(/4) 3 +.e j(/4) 4 +.e j(/4) 5 +.e j(/4) e j(/4)7 (/8)[ + 4/4 + /4 + 3/4 + 4/4 + 5/4 + 6/4 + 47/4 (/8)[ ].777 x[] (/8)[.e j(/4) + 4.e j(/4) +.e j(/4) +.e j(/4) 3 +.e j(/4) 4 +.e j(/4) 5 +.e j(/4) e j(/4)7 (/8)[ + 4/4 + 4/4 + 6/4 + 8/4 + /4 + /4 + 44/4 (/8)[].