EE 451 - HW 2

Solutions

1.

Problem 2.2

(a)

$$y[n] = x[n] * h[n] = \sum_{k=-\infty}^{\infty}h[k] x[n-k]$$

h[k] = 0 for k < N₀ and for k > N₁ so we only have to sum over these indices:

$$y[n] = x[n] * h[n] = \sum_{k=N_0}^{N_1}h[k] x[n-k]$$

We can change variables of summation by letting l = n - k, or k = n-l:

$$y[n] = x[n] * h[n] = \sum_{l=n-N_0}^{n-N_1}h[n-l] x[l]$$

Since x[l] = 0 for l<N₂, there will be no nozero x[l] in the sum if n-N₀ < N₂, or n < N₀ + N₂. Similarly, since x[l] = 0 for l>N₃, there will be no nozero x[l] in the sum if n-N₁ > N₃, or n < N₁ + N₃. Thus, y[n] will be zero for n < N₀+N₂ and for n > N₁ + N₃, and N₄ = N₀ + N₂, and N₅ = N₁ + N₃.

(b)

N = N₃ - N₂ + 1. M = N₁ - N₀ + 1. Nonzero points of y[n] go from N₄ to N₅, which is N₅ - N₄ + 1 long, or

N₅ - N₄ + 1 = (N₁ + N₃) - (N₀ + N₂) + 1 = (N₁ - N₀ +1) + (N₃ - N₂ +1) -1 = N + M - 1

2.

Problem 2.5

(a)

The characteristic equation is p² - 5 p + 6 = 0. This has roots p₁ = 2, p₂ = 3. Thus, the homogenous equation is

y_h[n] = C₁ 2ⁿ u[n] + C₂ 3ⁿ u[n]

(b)

h[n] = C₁ p₁ⁿ u[n] + C₂ p₂ⁿ u[n] = C₁ 2ⁿ u[n] + C₂ 3ⁿ u[n]

Thus h[0] = C₁ + C₂ and h[1] = 2 C₁ + 3 C₂. You can also find h[n] from the difference equation, where $x[n] = \delta[n]$:

$$h[0] = 5 h[-1] - 6 h[-2] + 2 \delta[-1] = 0$$

$$h[1] = 5 h[0] - 6 h[-1] + 2 \delta[0] = 2$$

Thus,

C₁ + C₂ = 0

2 C₁ + 3 C₂ = 2

This has the solution C₁ = -2 and C₂ = 2, so h[n] = -2 (2)ⁿ u[n] + 2 (3)ⁿ u[n].

(c)

The step response will be

g[n] = C₁ p₁ⁿ u[n] + C₂ p₂ⁿ u[n] + D u[n] = C₁ 2ⁿ u[n] + C₂ 3ⁿ u[n] + D u[n]

Thus, as in (b) g[0] = C₁ + C₂ + D, g[1] = 2 C₁ + 3 C₂ + D, and g[2] = 4 C₁ + 9 C₂ + D.

Also,

g[0] = 5 g[-1] - 6 g[-2] + 2 u[-1] = 0

g[1] = 5 g[0] - 6 g[-1] + 2 u[0] = 2

g[2] = 5 g[1] - 6 g[0] + 2 u[1] = 12

and

C₁ + C₂ + D = 0

2 C₁ + 3 C₂ + D = 2

4 C₁ + 9 C₂ + D = 12

This gives C₁ = -4, C₂ = 3, and D = 1, so

g[n] = -4 (2)ⁿ u[n] + 3 (3)ⁿ u[n] + u[n]

(d)

n = 0:20;
h = -2 * 2.^n + 2 * 3.^n;
subplot(211)
stem(n,h)
g = -4 * 2.^n + 3 * 3.^n + (n>=0);
subplot(212)
stem(n,g)

3.

Problem 2.18. An LTI system is causal if its impulse response is 0 for n<0.

(a)

Causal - h[n] turns on at n=0.

(b)

Causal - h[n] turns on at n=1.

(c)

Non-causal - e.g., h[-1] = 1/2

(d)

Non-causal - e.g., h[-1] = 1

(e)

Non-causal - e.g., h[-1] = 1/3

4.

No solution 4.

5.

Since x[n] = u[n] - u[n-10], all values of x[n] are zero except for n = 0 ... 9:

$$y[n] = x[n] * h[n] = \sum_{k=-\infty}^{\infty}x[k] h[n-k] = \sum_{k=0}^{9} h[n-k] = \sum_{k=0}^{9} (0.9)^{(n-k)} u[n-k]$$

For n<0, u[n-k] = 0 for all k (since k only goes from 0 to 9), so y[n] = 0 for n<0

For $0 \le n \le 9$, u[n-k] = 1 for k up to n in the sum, so

$$y[n] = \sum_{k=0}^{n} (0.9)^{(n-k)} = (0.9)^n \sum_{k=0}^{n} ... ...{-1})^{n+1}}{1-(0.9)^{-1}} = 10 \left[ 1 - (0.9)^{n+1} \right]$$

For n>9, u[n-k] = 1 for all k in the sum (since k only goes up to 9), so

$$y[n] = \sum_{k=0}^{9} (0.9)^{(n-k)} = (0.9)^n \sum_{k=0}^{9} ... ...}{1-(0.9)^{-1}} = 10 (0.9)^{n-9} \left[ 1 - (0.9)^{10} \right]$$

Here is a MATLAB program to plot the output:

   n=-5:50;                            
   y=zeros(size(n));                   
   ii=find((n>=0)&(n<=9));             
   y(ii)=10*(1-0.9.^n(ii));            
   ii=find(n>9);                       
   y(ii)=10*0.9.^(n(ii)-9)*(1-0.9^10);
   stem(n,y)

6.

(a)

$h[n] = h[n-1] + h[n-2] + \delta[n-1]$. Using this formula, $h[0] = h[-1] + h[-2] + \delta[-1] = 0$, and $h[1] = h[0] + h[-1] + \delta[0] = 1$. For n>1, h[n] = h[n-1] + h[n-2]. Thus, the impulse response of the difference equation gives the Fibonacci series.

(b)

The characteristic equation is p² - p - 1 = 0. This has the roots $p_{1,2} = (1 \pm \sqrt{5})/2$. Thus, y_h[n] = C₁ p₁ⁿ u[n] + C₂ p₂ⁿ u[n]. Since N = 2 and M = 1, there are no $\delta$s in the impulse response, and h[n] = C₁ p₁ⁿ u[n] + C₂ p₂ⁿ u[n]. Therefore

h[0] = C₁ + C₂ = 0

h[1] = C₁ p₁ + C₂ p₂ = 1

This gives $C_1 = \sqrt{5}$, and $C_2 = -\sqrt{5}$, and

$$h[n] = \frac{\sqrt{5}}{5} \left[ \frac{1 + \sqrt{5}}{2} \righ... ...\frac{\sqrt{5}}{5} \left[ \frac{1 - \sqrt{5}}{2} \right]^n u[n]$$