bevans@ece.utexas.edu
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
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where "oo" means infinity and Nh is the length of x[n] in samples.
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where "oo" means infinity and Nh is the length of x[n] in samples.
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Since the LTI system has a finite impulse response from n = 0 to n = Nh - 1,
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we can compute the summation for k = 0 to k = Nh - 1:
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we can compute the summation for k = 0 to k = Nh - 1:
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y[n] = h[0] x[n] + h[1] x[n-1] + h[2] x[n-2] + ... + h[Nh - 1] x[n - (Nh - 1)]We can substitute the values for h[n] = 1 for n = 0, 1, ..., Nh - 1,
y[n] = x[n] + x[n-1] + x[n-2] + ... + x[n - (Nh - 1)]
y[0] = x[0] + x[-1] + x[-2] + ... + x[-(Nh - 1)]The initial conditions are x[-1], x[-2], ..., x[-(Nh - 1)]. They must be set to zero as necessary conditions for LTI system properties to hold.
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:
Nmin = min(Nh, Nx)
Nmax = max(Nh, Nx)
conv command in MATLAB.
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:
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y[0] = 0.9 y[-1] + x[0]
y[1] = 0.9 y[0] + x[1]
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y[0] = 0.9 y[-1] + 0.1 x[0]
y[1] = 0.9 y[0] + 0.1 x[1]
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No initial conditions have arisen other than y[-1].
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
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)
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I 4
C
Y(z) = ----------------------- for |z| > max{ |a|, |b| }
-1 -1
(1 - a z ) (1 - b z )
E 4
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We can compute the inverse z-transform by rewriting the expression
as a sum of first-order denominators:
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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.
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D 4
2
C z A0 A1
Y(z) = --------------- = ----- + -----
(z - a) (z - b) z - a z - b
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z - z0
H(z) = C ------ for |z| > |p0|
z - p0
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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)
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D 9
where C is a constant.
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where C is a non-zero constant.
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E 6
bevans@ece.utexas.edu
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