Give step-by-step explanatiom and final Answer Which simplifies
t
o
:
0
.
8
y
(
n
)
+
0
.
1
y
(
n
-
3
)
-
0
.
3
y
(
n
-
5
)
-
0
.
4
y
(
n
-
2
)
-
0
.
5
y
(
n
-
1
)
=
x
(
n
)
+
x
(
n
+
1
)
This
i
s

Question

Give step-by-step explanatiom and final Answer Which simplifies 
t
o
:
0
.
8
y
(
n
)
+
0
.
1
y
(
n
-
3
)
-
0
.
3
y
(
n
-
5
)
-
0
.
4
y
(
n
-
2
)
-
0
.
5
y
(
n
-
1
)
=
x
(
n
)
+
x
(
n
+
1
)
This 
i
s
 ﻿the final form 
o
f
 ﻿the difference equation 
t
o
 
b
e
 ﻿used for solving parts 
(
a
)
 ﻿through 
(
g
)
.
a
.
 ﻿Determine the impulse response 
y
(
n
)
d
u
e
 
t
o
 ﻿the impulse sequence 
x
(
n
)
=
δ
(
n
)
b
.
 ﻿Determine system response 
y
(
n
)
d
u
e
 
t
o
 ﻿the unit step function excitation, where 
u
(
n
)
=
1
 ﻿for 
n
≥
0
c
.
 ﻿Find the transfer function 
H
(
z
)
d
.
 ﻿Plot the poles and zeros 
o
n
 ﻿the 
z
-
plane with the unit circle
e
.
 ﻿Determine the stability 
o
f
 ﻿the system from the pole
-
zero plot
f
.
 ﻿Calculate the amplitude 
(
m
a
g
n
i
t
u
d
e
)
 ﻿response 
o
f
 
H
(
z
)
g
.
 ﻿Calculate the phase response 
o
f
 
H
(
z
)
solve the question and plot 
i
f
 
w
e
 ﻿need

VIPSolutions · Accepted Answer

Let's carefully solve each part of the question step-by-step.

---

## **Given Difference Equation**

$$
0.8y(n) + 0.1y(n-3) - 0.3y(n-5) - 0.4y(n-2) - 0.5y(n-1) = x(n) + x(n+1)
$$

Let's rewrite it in standard form:

$$
0.8y(n) - 0.5y(n-1) - 0.4y(n-2) + 0.1y(n-3) - 0.3y(n-5) = x(n) + x(n+1)
$$

---

## **a. Impulse Response $ y(n) $ for $ x(n) = \delta(n) $ (Impulse Sequence)**

### **Step 1: Initial Conditions**

Since the system is causal, assume all $ y(k) = 0 $ for $ k < 0 $.

### **Step 2: Compute for each $ n $**

#### For $ n = 0 $:

- $ x(0) = 1 $, $ x(1) = 0 $
- All $ y $ terms with negative indices are zero.

$$
0.8y(0) = 1 + 0 \implies y(0) = \frac{1}{0.8} = 1.25
$$

#### For $ n = 1 $:

- $ x(1) = 0 $, $ x(2) = 0 $
- $ y(0) = 1.25 $, $ y(-1) = 0 $

$$
0.8y(1) - 0.5y(0) = 0
\implies 0.8y(1) = 0.5 \times 1.25 = 0.625
\implies y(1) = \frac{0.625}{0.8} = 0.78125
$$

#### For $ n = 2 $:

- $ x(2) = 0 $, $ x(3) = 0 $
- $ y(0) = 1.25 $, $ y(1) = 0.78125 $, $ y(-1) = 0 $

$$
0.8y(2) - 0.5y(1) - 0.4y(0) = 0
$$
$$
0.8y(2) = 0.5 \times 0.78125 + 0.4 \times 1.25 = 0.390625 + 0.5 = 0.890625
\implies y(2) = \frac{0.890625}{0.8} = 1.11328125
$$

#### For $ n = 3 $:

- $ y(3) $, $ y(2) = 1.11328125 $, $ y(1) = 0.78125 $, $ y(0) = 1.25 $, $ y(-2) = 0 $

$$
0.8y(3) - 0.5y(2) - 0.4y(1) + 0.1y(0) = 0
$$
$$
0.8y(3) = 0.5 \times 1.11328125 + 0.4 \times 0.78125 - 0.1 \times 1.25
$$
$$
= 0.556640625 + 0.3125 - 0.125 = 0.744140625
\implies y(3) = \frac{0.744140625}{0.8} = 0.93017578125
$$

#### For $ n = 4 $:

- $ y(4) $, $ y(3) = 0.93017578 $, $ y(2) = 1.11328125 $, $ y(1) = 0.78125 $, $ y(0) = 1.25 $, $ y(-1) = 0 $

$$
0.8y(4) - 0.5y(3) - 0.4y(2) + 0.1y(1) = 0
$$
$$
0.8y(4) = 0.5 \times 0.93017578 + 0.4 \times 1.11328125 - 0.1 \times 0.78125
$$
$$
= 0.46508789 + 0.4453125 - 0.078125 = 0.83227539
\implies y(4) = \frac{0.83227539}{0.8} = 1.04034424
$$

#### For $ n = 5 $:

- $ y(5) $, $ y(4) $, $ y(3) $, $ y(2) $, $ y(0) = 1.25 $, $ y(-1) = 0 $

$$
0.8y(5) - 0.5y(4) - 0.4y(3) + 0.1y(2) - 0.3y(0) = 0
$$
$$
0.8y(5) = 0.5 \times 1.04034424 + 0.4 \times 0.93017578 - 0.1 \times 1.11328125 + 0.3 \times 1.25
$$
$$
= 0.52017212 + 0.37207031 - 0.111328125 + 0.375
$$
$$
= 0.52017212 + 0.37207031 = 0.89224243
$$
$$
0.89224243 - 0.111328125 = 0.780914305
$$
$$
0.780914305 + 0.375 = 1.1559143
$$
$$
y(5) = \frac{1.1559143}{0.8} = 1.4448929
$$

### **Impulse Response (first six samples):**

$$
\boxed{
y(0) = 1.25, \quad y(1) = 0.78125, \quad y(2) = 1.11328, \quad y(3) = 0.93018, \quad y(4) = 1.04034, \quad y(5) = 1.44489
}
$$

---

## **b. Step Response $ y(n) $ for $ x(n) = u(n) $**

$ x(n) = 1 $ for $ n \geq 0 $.

#### For $ n = 0 $:

$$
0.8y(0) = 1 + 1 = 2 \implies y(0) = \frac{2}{0.8} = 2.5
$$

#### For $ n = 1 $:

$$
0.8y(1) - 0.5y(0) = 1 + 1 = 2
$$
$$
0.8y(1) = 2 + 0.5 \times 2.5 = 2 + 1.25 = 3.25
\implies y(1) = \frac{3.25}{0.8} = 4.0625
$$

#### For $ n = 2 $:

$$
0.8y(2) - 0.5y(1) - 0.4y(0) = 1 + 1 = 2
$$
$$
0.8y(2) = 2 + 0.5 \times 4.0625 + 0.4 \times 2.5 = 2 + 2.03125 + 1 = 5.03125
$$
$$
y(2) = \frac{5.03125}{0.8} = 6.2890625
$$

#### For $ n = 3 $:

$$
0.8y(3) - 0.5y(2) - 0.4y(1) + 0.1y(0) = 2
$$
$$
0.5y(2) = 0.5 \times 6.2890625 = 3.14453125
$$
$$
0.4y(1) = 0.4 \times 4.0625 = 1.625
$$
$$
0.1y(0) = 0.25
$$
$$
0.8y(3) = 2 + 3.14453125 + 1.625 - 0.25 = 6.51953125
$$
$$
y(3) = \frac{6.51953125}{0.8} = 8.14941406
$$

#### For $ n = 4 $:

$$
0.8y(4) - 0.5y(3) - 0.4y(2) + 0.1y(1) = 2
$$
$$
0.5y(3) = 0.5 \times 8.14941406 = 4.07470703
$$
$$
0.4y(2) = 0.4 \times 6.2890625 = 2.515625
$$
$$
0.1y(1) = 0.1 \times 4.0625 = 0.40625
$$
$$
0.8y(4) = 2 + 4.07470703 + 2.515625 - 0.40625 = 8.18408203
$$
$$
y(4) = \frac{8.18408203}{0.8} = 10.2301025
$$

---

## **c. Transfer Function $ H(z) $**

Take z-transform (assume zero initial conditions):

$$
0.8Y(z) - 0.5z^{-1}Y(z) - 0.4z^{-2}Y(z) + 0.1z^{-3}Y(z) - 0.3z^{-5}Y(z) = X(z) + zX(z)
$$

$$
Y(z)[0.8 - 0.5z^{-1} - 0.4z^{-2} + 0.1z^{-3} - 0.3z^{-5}] = X(z)[1 + z]
$$

$$
H(z) = \frac{Y(z)}{X(z)} = \frac{1 + z}{0.8 - 0.5z^{-1} - 0.4z^{-2} + 0.1z^{-3} - 0.3z^{-5}}
$$

Multiply numerator and denominator by $ z^5 $:

$$
H(z) = \frac{z^6 + z^5}{0.8z^5 - 0.5z^4 - 0.4z^3 + 0.1z^2 - 0.3}
$$

---

## **d. Poles and Zeros on z-plane**

### **Zeros:**
Where numerator is zero:
$$
z^6 + z^5 = 0 \implies z^5(z + 1) = 0
$$
- Five zeros at $ z = 0 $
- One zero at $ z = -1 $

### **Poles:**
Solve denominator:
$$
0.8z^5 - 0.5z^4 - 0.4z^3 + 0.1z^2 - 0.3 = 0
$$
This is a 5th-order equation, roots can be found numerically.

---

### **Python code for plotting:**

```python
import numpy as np
import matplotlib.pyplot as plt
from numpy.polynomial import Polynomial

# Numerator and denominator coefficients (highest power first)
num = [1, 1, 0, 0, 0, 0, 0]   # z^6 + z^5
den = [0.8, -0.5, -0.4, 0.1, 0, 0, -0.3]  # 0.8z^5 - 0.5z^4 - 0.4z^3 + 0.1z^2 - 0.3

zeros = np.roots(num)
poles = np.roots(den)

plt.figure(figsi

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