h26005 s 00054/00003/00200 d D 1.13 24/10/24 12:51:55 bevans 13 12 c Updated e s 00003/00000/00200 d D 1.12 23/11/02 13:06:04 bevans 12 11 c Up e s 00015/00072/00185 d D 1.11 23/11/01 15:04:09 bevans 11 10 c Updated e s 00032/00000/00225 d D 1.10 21/11/12 07:27:27 bevans 10 9 c Updated e s 00017/00004/00208 d D 1.9 21/11/11 13:33:26 bevans 9 8 c Updated e s 00001/00001/00211 d D 1.8 21/11/11 10:46:12 bevans 8 7 c Updated e s 00035/00000/00177 d D 1.7 21/11/11 10:35:18 bevans 7 6 c Updated e s 00003/00000/00174 d D 1.6 21/11/10 23:17:47 bevans 6 5 c Updted e s 00001/00001/00173 d D 1.5 21/11/10 22:13:00 bevans 5 4 c Updated e s 00034/00022/00140 d D 1.4 21/11/10 17:41:44 bevans 4 3 c Updated e s 00025/00000/00137 d D 1.3 21/11/10 16:53:24 bevans 3 2 c Updated e s 00021/00011/00116 d D 1.2 21/11/10 16:45:40 bevans 2 1 c Updated e s 00127/00000/00000 d D 1.1 21/11/09 21:27:33 bevans 1 0 c date and time created 21/11/09 21:27:33 by bevans e u U f i f e 0 t T I 1 D 2 EE313 Linear Systems & Signals - Homework 6 Hints E 2 I 2 EE313 Linear Systems & Signals - Homework 7 Hints E 2 D 2

EE313 Linear Systems & Signals - Homework 6 Hints

E 2 I 2

EE313 Linear Systems & Signals - Homework 7 Hints

E 2 Homework #7 assignment

Problem 7.1(a). E 13 I 13
Problem 7.1. This problem is the same as problem 6.3 in fall 2021.
In homework problem 6.3(a), a linear time-invariant (LTI) system with input signal x[n] and output signal y[n] is described by
y[n] = x[n] + x[n - 1] for n ≥ 0
As a necessary condition for LTI properties to hold, the LTI system must be at rest. This means that the initial condition x[-1] = 0.
The LTI system is a version of the two-point averaging filter in which the coefficients of 1/2 and 1/2 have been scaled by 2.
1. The impulse response of a system is the system response (output) when the input is an impulse signal. Denoting the impulse response as h[n] and letting the input signal be the impulse signal d[n]
  h[n] = d[n] + d[n - 1] for n ≥ 0
2. The transfer function in the z-domain is the z-transform of the impulse response h[n] :
```
         oo          -n    1         -n                -1
H(z) =   Sum   h[n] z   = Sum  h[n] z   = h[0] + h[1] z  
       n = -oo            n=0                            
```
3. To find the roots of the numerator (zeros) and the roots of the denominator (poles) of H(z),
```
            -1
H(z) = 1 + z  
```
  Zero is at z + 1 = 0 or z = -1.
  Pole is at z = 0.
4. The MATLAB code to plot the poles and zeros in the complex z plane is
```
h = [1 1];
zplane(h);
```
  Note that zplane has the same syntax as freqz.
Homework problems 6.3(b) and 6.3(c) are very similar.
Problem 7.2(a). E 13
- Please see the handout Design Averaging Filters
  E 12
- i. "Difference equation" means the input-output relationship between y[n] and x[n]. Because the system is linear and time-invariant (LTI) ,
```
                       oo                 Nh-1
y[n] = h[n] * x[n] =  Sum  h[k] x[n-k] =  Sum  h[k] x[n-k] 
                     k=-oo                k=0
```
  D 2 where "oo" means infinity and Nh is the length of x[n] in samples. E 2 I 2 where "oo" means infinity and N_h is the length of x[n] in samples. E 2 Since the LTI system has a finite impulse response from n = 0 to n = N_h - 1, D 2 we can compute the summation for k = 0 to k = Nh - 1: E 2 I 2 we can compute the summation for k = 0 to k = N_h - 1: E 2
```
y[n] = h[0] x[n] + h[1] x[n-1] + h[2] x[n-2] + ... + h[N_h - 1] x[n - (N_h - 1)]
```
  We can substitute the values for h[n] = 1 for n = 0, 1, ..., N_h - 1,
```
y[n] = x[n] + x[n-1] + x[n-2] + ... + x[n - (N_h - 1)]
```
- ii. Initial conditions can be discovered by computing the first several values of y[n] for n ≥ 0:
```
y[0] = x[0] + x[-1] + x[-2] + ... + x[-(N_h - 1)]
```
  The initial conditions are x[-1], x[-2], ..., x[-(N_h - 1)]. They must be set to zero as necessary conditions for LTI system properties to hold.
- iii. In this part, you'll develop a piecewise representation for the convolution result. The convolution of two finite-length signal will provide a piecewise representation of five intervals: no overlap, partial overlap, complete overlap, partial overlap, and no overlap.
  In discrete time, the convolution of two finite-length signals h[n] and x[n] produces a result that is as long as the sum of the lengths of h[n] and x[n] minus one.
  The convolution of two rectangular pulses of equal length will produce a triangle whose width is the twice the width of the rectangular pulse minus one. For more information, please see Handout E Convolving Two Rectangular Pulses.
  The convolution of two rectangular pulses of different lengths will produce a trapezoid whose duration is the sum of the durations of the two pulses minus one. The trapezoid corresponds to the five cases of convolving two finite-length signals when flipping-and-sliding one signal across the other: no overlap, partial overlap, complete overlap, partial overlap, and no overlap. The plateau of the trapezoid corresponds to the complete overlap case.
  Using the following relationships might be helpful in this problem:
  N_min = min(N_h, N_x)
  N_max = max(N_h, N_x)
- iv. You can use the conv command in MATLAB.
- v. Convolution of signals h[n] and x[n] in the discrete-time domain becauses the product of their z-transforms H(z) and X(z) in the z-domain. After computing Y(z) = H(z) X(z), use the inverse z-transform to find y[n].
D 13
Problem 7.1(b). E 13 I 13
Problem 7.2(b). E 13
- i. The input-output relationship for the system is
  D 2 y[n] = 0.9 y[n - 1] + x[n] E 2 I 2 y[n] = 0.9 y[n - 1] + 0.1 x[n] E 2
  for n ≥ 0. We can find the initial values by computing the first output values:
  D 2 y[0] = 0.9 y[-1] + x[0]
  y[1] = 0.9 y[0] + x[1] E 2 I 2 y[0] = 0.9 y[-1] + 0.1 x[0]
  y[1] = 0.9 y[0] + 0.1 x[1] E 2
  No initial conditions have arisen other than y[-1].
- ii. To compute the impulse response h[n], we let the input be an impulse d[n]:
  D 2 h[n] = 0.9 h[n - 1] + d[n] E 2 I 2 h[n] = 0.9 h[n - 1] + 0.1 d[n] E 2
  D 2 for n ≥ 0. We could compute output values manually and infer the impulse response, or use the z-transform. E 2 I 2 for n ≥ 0. We could compute output values manually and infer the impulse response as we did in LTI Example #2 on Lecture Slide 11-4, or use the z-transform. E 2
- iii. This problem involves convolving two causal exponential sequences. Handout F Convolution of Two Causal Exponential Sequences might be helpful. I 2
- iv. Here, h[n] = C aⁿ u[n] and x[n] = bⁿ u[n]. In the discrete-time domain, the output y[n] = h[n] * x[n]. I 4
  E 4 In the z-domain, Y(z) = H(z) X(z). D 4 The z-transform of a signal bⁿ u[n] is 1 / (1 - b z^-1) for |z| > |b|. E 4 I 4 The z-transform of x[n] = bⁿ u[n] is X(z) = 1 / (1 - b z^-1) for |z| > |b|. E 4 That is, the region of convergence is |z| > |b|. I 3
  I 4 For Y(z), the valid values of z must be those that are valid for H(z) and X(z). Hence, the region of convergence for Y(z) is the intersection of the regions of convergence for H(z) and X(z): E 4
```
D 4
                                          2
                 C                     C z
Y(z) = ----------------------- = ---------------
               -1          -1
       (1 - a z  ) (1 - b z  )   (z - a) (z - b)
E 4
I 4
                 C              
Y(z) = -----------------------   for |z| > max{ |a|, |b| }
               -1          -1 
       (1 - a z  ) (1 - b z  ) 
E 4
```
  D 4 We can compute the inverse z-transform by rewriting the expression as a sum of first-order denominators: E 4 I 4 We can use partial fractions decomposition to express the transfer function as a sum of two first-order terms and then apply the inverse z-transform to each term. See Example 8-10 on page 219.
D 5
Problem 7.2. E 5 I 5 D 13
Problem 7.2. E 13 I 13
Problem 7.3. E 13 E 5
- i. For Pole-Zero Plot #1, I would estimate the pole location ("X") to be at p0 = -0.9 and the zero location ("O") to be at z0 = 1. That would give a transfer function in the z-domain of E 4
```
D 4
                2
             C z          A0      A1
Y(z) = --------------- = ----- + -----
       (z - a) (z - b)   z - a   z - b
E 4
I 4
         z - z0
H(z) = C ------   for |z| > |p0|
         z - p0
E 4
```
  D 4 We put the sum of the two first-order terms under a common denominator to solve for A0 and A1:
```
                2
             C z           A0      A1    (A0 + A1) z - (A0 b + A1 a)
Y(z) = --------------- = ----- + ----- = ---------------------------
       (z - a) (z - b)   z - a   z - b         (z - a) (z - b)
```
  E 4 I 4 D 9 where C is a constant. E 9 I 9 where C is a non-zero constant. E 9
- ii. The question should have read "why can we convert from the transfer function in the z-domain to a frequency response directly?". I 9 Here's more information about converting from z to frequency domain E 9
  I 6
- iii. Please provide the plots of the magnitude response to validate the transfer function in the z-domain. I 9 Here's example Matlab code bandstopfilter.m to design a bandstop filter and plot its magnitude response. E 9
  E 6
- iv. The magnitude response in (A) passes low frequencies around 0 rad/sample and attenuates high frequencies above say pi/4 rad/samples. Frequencies at -pi and pi rad/samples are zeroed out. This is a lowpass filter. I 9 Here's an example of a bandstop filter. E 9 E 4 E 3 E 2
I 7 D 11
Problem 7.3. E 11 I 11 Connecting filter poles/zeros to its frequency selectivity: Please see lecture slides 11-6 through 11-11 and watch the recording from our lecture on Oct. 31, 2023, from 1:42 to 39:03, which is available on Canvas. I have another recording of the same demos on YouTube video in spring 2014 for the Real-Time Digital Signal Processing Lab course from the 1:29 to 22:25 and from 43:01 to the end (50:51). Takeaways from either videorecording: E 11
- Assume that we can observe the system for all time -oo < t < oo.
- Linearity. When checking each system for linearity, we can use the quick test of input D 9 signal of 0 for all time. E 9 I 9 signal of 0 for all time, which is a by-product of the homogeneity property when the input signal is scaled by a = 0. E 9 If the output is not zero for all time, then the system is not linear. If the output is zero for all time, then we'll have to then apply the mathematical definitions for homogeneity and additivity.
  - Systems (a), (b), and (c) are not linear.
  - System (d) is linear. E 9 I 9
  - Systems (a), (b), and (c) are not linear. All fail the all-zero input signal test.
  - System (d) is linear. E 9
- Time-Invariance. For time-invariant system, shift of the input signal by any real-valued tau causes same shift in output signal, i.e. x(t- tau) means y(t - tau) for all tau.
  - System (a) is time-invariant.
  - Systems (b), (c), and (d) are time-varying.
- Stability. A stable system will always produce a bounded amplitude output signal when given a bounded amplitude input signal.
  - All four systems are stable.
- Causality. A causal system depends only on current and previous input values and/or previous output values to compute an output value.
  - Systems (a) and (d) are not causal. D 8
  - Systems (b) and (d) are causal. E 8 I 8
  - Systems (b) and (c) are causal. I 9
  - For System (d), we can compute different output values.
    - When t = 0, y(0) = ( x(0) + x(0) ) / 2 = x(0). Causal.
    - When t = 2, y(2) = ( x(2) + x(-2) ) / 2. Causal.
    - When t = -2, y(-2) = ( x(-2) + x(2) ) / 2. Not Causal. Depends on future input value.
    E 9 E 8
  E 11 I 11
- When poles and zeros are separated in angle, the angles of the poles near the unit circle indicate the frequencies in the passband(s) and the angles of the zeros near or on the unit circle indicate the frequencies in the stopband(s).
- Poles must be inside the unit circle for bounded-input bounded-output (BIBO) stability. Please see lecture slides 11-12 and 11-13. E 11
I 10 D 11
Problem 7.4(a)
- Convolving two rectangular pulses of different widths gives a trapezoid whose width is the sum of the widths of the two rectangular pulses
- Let
  - T_min = min(T_h, T_x)
  - T_max = max(T_h, T_x)
  - T_y = T_h + T_x
- As we flip and slide one rectangular pulse against the other, partial overlap occurs from 0 to T_min seconds, complete overlap from T_min to T_max seconds, and partial overlap from T_max to T_y seconds.
- We can check the points at the boundaries between intervals for a sanity check. For T_h = 4 and T_x = 9, the transition point between partial overlap and complete overlap occurs at t = 4 seconds. At t = 4 seconds, the amplitude for partial overlap and complete overlap should be the same.
Problem 7.4(b)
- The integrator was operating before t = 0 seconds, but we aren't able to observe that. We are observing the system starting at t = 0 seconds.
- Since we are starting the integration at t = 0 seconds, there is ambiguity as to whether Dirac delta signal would be included at time 0. A better notation for the lower limit on the integration might have been 0− to indicate that the integration starts at time 0 before the impulse occurs. Clunky notation, but one that is used.
E 11 I 11 E 11 E 10 E 7

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