Solved Problems on Signals and Systems: Electrical Engineering 201

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Title

Question 1
Step 1: Ordering Pizza
The given figure shows the pizza ordering. Description of this step follows:
i) The guest asks for the pizza menu and the host provides the pizza menu to them and by
following the menu the guest selects the pizza and place the order.
ii) The guest has to look at the telephone number of host and then dials that number on the
telephone. The telephone line is connected and send tones to that number.
iii) Host confirms and allocates the order with the order clerk and then acknowledge the order
with the guest.
iv) The guest has to wait for the order.
v) After receiving the order by order clerk, he has to ask for payment confirmation and then after
approval he places the order to the pizza cook.
Step 2: Delivering of Pizza
i) After receiving the order the order cook started preparing the pizza according to the order
received from guest and handles that order to order clerk for delivery.
ii) Order clerk has the responsibility to pack the pizza with the receipt and put that in the delivery
van.
iii) The delivery van delivers all the order from order clerk to the host and delivery van travels by
the road as transmitting line.
iv) Now after receiving the orders host confirms all the payments to the order clerk and then
acknowledge delivery.
v) Host delivers all the pizza orders to the guests respectively.

Question 2
a)
b)

Question 3
(1)
From the given figure:
Maximum Amplitude (Am) = 15 …… (1)
Time (T) = 3 sec.
 frequency (f) = (1/T) = 1/3
 f = 0.333 Hz ...... (2)
x (t) = Am Sin (t + ) (here  = phase)
Here t = 0
 x (t) = 0
 Am Sin () = 0
  = 0
So phase  = 0• ….. (3)

(2)
From the given Figure:
Maximum Amplitude (Am) = 4 …... (1)
Time (T) = 6.5 sec.
 frequency (f) = (1/T) = 1/6.5
 f = 0.1538 Hz ...... (2)
Here also t = 0
 x (t) = 0
 Am Sin () = 0
  = 0
So phase  = 0• ….. (3)

(3)
From the given Figure:
Maximum Amplitude (Am) = 7.5 …... (1)
Time (T) = 2.25 sec.
 frequency (f) = (1/T) = 1/2.25
 f = 0.444 Hz ...... (2)
Here at t = 0; Am is max
 Am Sin (t + ) = Am
 Sin () = 1
  = 90
So phase  = 90• ….. (3)

Question 4
Normal Waveform: ASin (2п(f)t)
Here A = Amplitude
f = frequency
So Time (T) = 1/f
a) 10 Sin(2п(100)t)
 A = 10
 f = 100 Hz
 T = 1/f = 1/100 = 0.015 sec.
 Phase  = 0
b) 20 Sin(2п(30)t)

 A = 20
 f = 30 Hz
 T = 1/f = 1/30 = 0.33 sec.
 Phase  = 0
c) 5 Sin(500пt + 180)
 A = 5
 f = 500/2 = 250 Hz
 T = 1/f = 1/250 = 0.04 sec.
 Phase  = 180• (Phase shift to the left plot)

d) 8 Sin(400пt + 270)
 A = 8
 f = 400/2 = 200 Hz
 T = 1/f = 1/200 = 0.005 sec.
 Phase  = 270•

Question 5
a) Calculating the source rate R:
Given:
Pixels = 480 * 500
Per second sent pictures = 30
So pixel/sec = (480 * 500) * 30 = 7200000
Each pixel can take 32 intensity values:
So log2 (32) = 5 (by logarithm rules)
 Source rate R = 7200000 * 5 (we got 5 bits/pixel)
 Source rate R = 36000000
 Source rate R = 36 * 106 mbps
b) Given Bandwidth B = 4.5 MHz
 B = 4.5 106 Hz
SN Ratiodb = 35 = 10 log10 (35)
 3162
C = B log2 (1 + SNR)
= 4.5 * 106 log2 (1 + 3162)
= 4.5 * 106 log2 (3163)
= 4.5 * 106 * 11.6272
= 52.3 * 106 bps

Question 6
Given:
frequency = 4GHz = 4 * 109 Hz
Satellite Distance from Earth = 35863 Km = 35.863 * 106 meter
So Free Space Loss:
= 20 log10 (4 * 109) + 20 log10 (35.863 * 106) – 147.56
= 195.57 db

Question 7
S (t) = 5 Sin(200пt) + Sin(600пt)
S (t) = 5 Sin(2пf1t) + Sin(2пf2t)
We can assume that we have three frequencies f1, f2, f3. All the signals are integer multiple of f0.
So f1 = n1f0
f2 = n2f0
f3 = n3f0
f0 = GCD (f1, f2, f3)
So the given signal f0 = GCD (100, 300) = 100 (which is fundamental frequency)
S (t) = 5 Sin (200пt) + Sin (600пt)
Fourier Transformation: 1/2i [δ (f - A) – δ (f + A)]
 5 * 1/2i [δ (f - 100) – δ (f + 100)] + 1/2i [δ (f - 300) – δ (f + 300)]
 1/2i [5 * [δ (f - 100) – δ (f + 100)] + 1/2i [δ (f - 300) – δ (f + 300)]]
B = fmax - fmin
= (f + 300) – (f – 100) = f + 300 – f + 100
So B = 400 Hz
Channel Capacity = 2 * Bandwidth * log2L
Here L = level number
For level 2:
Channel Capacity (bit rate) = 2 * 400 * log22 = 800 bps
For level 4:
Channel Capacity (bit rate) = 2 * 400 * log24 = 1600 bps
For level 8:
Channel Capacity (bit rate) = 2 * 400 * log28 = 2400 bps