NOISE IN COMMUNICATION SYSTEMS (ADC-UNIT-V)

A signal mainly be contaminated along the path by noise. Noise may be defined as any unwanted
introduction of energy into the desired signal. In radio receivers, noise may produce ‘hiss in the
loudspeaker output. Noise is random and unpredictable. Noise is produced both external and
internal to the system. External noise includes atmospheric noise (e.g. lightning), Galactic noise
(thermal radiation from cosmic bodies), and industrial noise (e.g. motors’ ignition). We can
minimize or eliminate external noise by proper system design. On the other hand internal noise is
generated inside the system. It is resulted due to random motion of charged particles in
conductors and electronic devices. With proper system design it can be minimized but never can
be eliminated. This is the major constraint in the ra1e of telecommunications.

Noise Figure
The Noise figure is the amount of noise power added by the electronic circuitry in the receiver to
the thermal noise power from the input of the receiver. The thermal noise at the input to the
receiver passes through to the demodulator. This noise is present in the receive channel and
cannot be removed. The noise figure of circuits in the receiver such as amplifiers and mixers,
adds additional noise to the receiver. This raises the noise power to demodulator.
𝐼𝑛𝑝𝑢𝑡 𝑆𝑖𝑔𝑛𝑎𝑙 𝑡𝑜 𝑁𝑜𝑖𝑠𝑒 𝑅𝑎𝑡𝑖𝑜 (𝑆𝑁𝑅)𝑖
𝑁𝑜𝑖𝑠𝑒 𝐹𝑖𝑔𝑢𝑟𝑒 (𝑁𝐹) = =
𝑂𝑢𝑡𝑝𝑢𝑡 𝑆𝑖𝑔𝑛𝑎𝑙 𝑡𝑜 𝑁𝑜𝑖𝑠𝑒 𝑅𝑎𝑡𝑖𝑜 (𝑆𝑁𝑅)0
And figure of merit is reciprocal of noise figure
𝑂𝑢𝑡𝑝𝑢𝑡 𝑆𝑖𝑔𝑛𝑎𝑙 𝑡𝑜 𝑁𝑜𝑖𝑠𝑒 𝑅𝑎𝑡𝑖𝑜 (𝑆𝑁𝑅)0
𝐹𝑖𝑔𝑢𝑟 𝑜𝑓 𝑀𝑒𝑟𝑖𝑡 (𝐹𝑀) = =
𝐼𝑛𝑝𝑢𝑡 𝑆𝑖𝑔𝑛𝑎𝑙 𝑡𝑜 𝑁𝑜𝑖𝑠𝑒 𝑅𝑎𝑡𝑖𝑜 (𝑆𝑁𝑅)𝑖

Noise in AM system (Envelope Detection):

Let the transmitted DSBSCAM signal be
𝑠(𝑡) = 𝐴𝑐 [1 + 𝑘𝑎 𝑚(𝑡)] cos(2𝜋𝑓𝑐 𝑡)
The received signal at the input of the receiver is added to the white noise and filtered in a BPF.
A filtered noise can be expressed in terms of its in-phase and quadrature components as
𝑛(𝑡) = 𝑛𝑐 (𝑡) cos(2𝜋𝑓𝑐 𝑡) − 𝑛𝑠 (𝑡) sin(2𝜋𝑓𝑐 𝑡)
Let us consider AM receiver as shown in the figure with input 𝑠(𝑡) + 𝑛(𝑡)
𝑦(𝑡) = 𝑠𝑖 (𝑡) + 𝑛𝑖 (𝑡)
Input
𝑠(𝑡) + 𝑛(𝑡) Envelope
BPF LPF 𝑣0 (𝑡) = 𝑠0 (𝑡) + 𝑛0 (𝑡)
Detector

After BPF the noise is bandpass noise. Then the output of BPF is sum of modulated signal and
bandpass noise with nc(t) as in-phase component and ns(t) as quadrature phase component.
Therefore the input of detector is
𝑦(𝑡) = 𝑠𝑖 (𝑡) + 𝑛𝑖 (𝑡)
= 𝐴𝑐 [1 + 𝑘𝑎 𝑚(𝑡)] cos(2𝜋𝑓𝑐 𝑡) + 𝑛𝑐 (𝑡) cos(2𝜋𝑓𝑐 𝑡) − 𝑛𝑠 (𝑡) sin(2𝜋𝑓𝑐 𝑡)
Input signal power is given by
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
2
𝐴𝑐 2 [1 + 𝑘𝑎 2̅̅̅̅̅̅̅̅
𝑚2 (𝑡)] 𝐴𝑐 2 [1 + 𝑘𝑎 2 𝑃𝑚 ]
𝑃𝑠𝑖 = {𝐴𝑐 [1 + 𝑘𝑎 𝑚(𝑡)] cos(2𝜋𝑓𝑐 𝑡)} = =
2 2
Where 𝑃𝑚 is the modulating power.
Input noise power is given by 𝐺𝑛i (𝑓)
𝑓𝑐 +𝑤

̅̅̅̅̅̅̅̅  
𝑃𝑛𝑖 = 𝑛 2
𝑖 (𝑡) = 2 ∫ 𝑑𝑓 = 2𝑊
2 2
𝑓𝑐 −𝑤
Input signal to noise ratio is given by
𝐴𝑐 2 [1+𝑘𝑎 2 𝑃𝑚 ] −𝑓𝑐 − 𝑤 −𝑓𝑐 + 𝑤 0 𝑓𝑐 + 𝑤
𝑓𝑐 − 𝑤
𝑃𝑠𝑖 2 𝐴𝑐 2 [1 + 𝑘𝑎 2 𝑃𝑚 ]
(𝑆𝑁𝑅)𝑖 = = =
𝑃𝑛𝑖 2𝑊 4𝑊
The received signal is demodulated by using envelope detector which will respond to the
envelope of the input. Let us write the input which is the sum of input signal and banpass noise
interms of a single sinusoid as follows
𝑦(𝑡) = 𝐴𝑐 [1 + 𝑘𝑎 𝑚(𝑡)] cos(2𝜋𝑓𝑐 𝑡) + 𝑛𝑐 (𝑡)cos(2𝜋𝑓𝑐 𝑡) − 𝑛𝑠 (𝑡) sin(2𝜋𝑓𝑐 𝑡)
= [𝐴𝑐 + 𝐴𝑐 𝑘𝑎 𝑚(𝑡) + 𝑛𝑐 (𝑡)] cos(2𝜋𝑓𝑐 𝑡) − 𝑛𝑠 (𝑡) sin(2𝜋𝑓𝑐 𝑡)

= 𝐸 cos(2𝜋𝑓𝑐 𝑡 + ∅)

Where

𝐸 = √{[𝐴𝑐 + 𝐴𝑐 𝑘𝑎 𝑚(𝑡) + 𝑛𝑐 (𝑡)]2 + [𝑛𝑠 (𝑡)]2 }

𝑛𝑠 (𝑡)
And ∅ = 𝑡𝑎𝑛−1 [[𝐴 ]
𝑐 +𝐴𝑐 𝑘𝑎 𝑚(𝑡)+𝑛𝑐 (𝑡)]

Therefore the output of the envelope detector is given by

𝐸 = √{[𝐴𝑐 + 𝐴𝑐 𝑘𝑎 𝑚(𝑡) + 𝑛𝑐 (𝑡)]2 + [𝑛𝑠 (𝑡)]2 }

When the noise is small compared to 𝐴𝑐 , then [𝑛𝑠 (𝑡)]2 can be neglected by comparing with
[𝐴𝑐 + 𝐴𝑐 𝑘𝑎 𝑚(𝑡) + 𝑛𝑐 (𝑡)]2 . Therefore the envelope is given by

𝐸 = 𝐴𝑐 + 𝐴𝑐 𝑘𝑎 𝑚(𝑡) + 𝑛𝑐 (𝑡)
When the dc term 𝐴𝑐 is removed by passing through a capacitor the output is given by
𝐸 = 𝐴𝑐 𝑘𝑎 𝑚(𝑡) + 𝑛𝑐 (𝑡)
= 𝑠0 (𝑡) + 𝑛0 (𝑡)
Therefore the output signal power is given by
𝑃𝑠0 = ̅̅̅̅̅̅̅̅
𝑠0 (𝑡)2 = ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
[𝐴𝑐 𝑘𝑎 𝑚(𝑡)]2 = 𝐴2𝑐 𝑘𝑎2̅̅̅̅̅̅̅̅
𝑚2 (𝑡) = 𝐴2𝑐 𝑘𝑎2 𝑃𝑚
Where 𝑃𝑚 is the modulating power.
And output noise power is given by
 ̅̅̅̅̅̅̅̅ ̅̅̅̅̅̅̅̅̅̅2 𝑤
0 (𝑡) = [𝑛𝑐 (𝑡)] = ∫−𝑤  𝑑𝑓 = 2𝑊
𝑃𝑛0 = 𝑛 2
𝐺𝑛c (𝑓)
Therefore output SNR is given by
𝑃𝑠0 𝐴2𝑐 𝑘2𝑎 𝑃𝑚 𝐴2𝑐 𝑘2𝑎 𝑃𝑚 𝑃𝑠𝑖 
(𝑆𝑁𝑅)0 = = = ∗
𝑃𝑛0 2𝑊 2𝑊 𝑃𝑠𝑖
𝐴2𝑐 𝑘2𝑎 𝑃𝑚 𝑃𝑠𝑖
= ∗
2𝑊 𝐴𝑐 [1+𝑘𝑎 2 𝑃𝑚 ]
2

2
𝑓
−𝑊 𝑊
𝑃𝑠𝑖 𝑘2𝑎 𝑃𝑚
= ∗
𝑊 1 + 𝑘𝑎 2 𝑃𝑚
𝐴2𝑐 𝑘2𝑎 𝑃𝑚
(𝑆𝑁𝑅)0 2𝑊 2𝑘2𝑎 𝑃𝑚
𝐹𝑖𝑔𝑢𝑟𝑒 𝑜𝑓 𝑀𝑒𝑟𝑖𝑡 = = =
(𝑆𝑁𝑅)𝑖 𝐴𝑐 2 [1+𝑘𝑎 2 𝑃𝑚 ] 1 + 𝑘𝑎 2 𝑃𝑚
4𝑊
Noise in DSBSC system:
Let the transmitted DSBSC signal is
𝑠(𝑡) = 𝐴𝑐 𝑚(𝑡) cos(2𝜋𝑓𝑐 𝑡)
The received signal at the input of the receiver is added to the white noise and filtered in a BPF.
A filtered noise can be expressed in terms of its in-phase and quadrature components as
𝑛(𝑡) = 𝑛𝑐 (𝑡) cos(2𝜋𝑓𝑐 𝑡) − 𝑛𝑠 (𝑡) sin(2𝜋𝑓𝑐 𝑡)
Let us consider DSBC receiver as shown in the figure with input 𝑠(𝑡) + 𝑛(𝑡)
𝑦(𝑡) = 𝑠𝑖 (𝑡) + 𝑛𝑖 (𝑡)
Input
𝑠(𝑡) + 𝑛(𝑡) Product
BPF LPF 𝑣0 (𝑡) = 𝑠0 (𝑡) + 𝑛0 (𝑡)
Modulator

cos(2𝜋𝑓𝑐 𝑡)
After BPF the noise is bandpass noise. Then the output of BPF is sum of modulated signal and
bandpass noise with nc(t) as in-phase component and ns(t) as quadrature phase component.
Therefore the input of detector is
𝑦(𝑡) = 𝑠𝑖 (𝑡) + 𝑛𝑖 (𝑡) = 𝐴𝑐 𝑚(𝑡) cos(2𝜋𝑓𝑐 𝑡) + 𝑛𝑐 (𝑡) cos(2𝜋𝑓𝑐 𝑡) − 𝑛𝑠 (𝑡) sin(2𝜋𝑓𝑐 𝑡)

Input signal power is given by

𝐴𝑐 2 ̅̅̅̅̅̅̅̅̅ 𝐴𝑐 2 𝐺𝑛i (𝑓)
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
𝑃𝑠𝑖 = [𝐴𝑐 𝑚(𝑡) cos(2𝜋𝑓𝑐 𝑡)] = 2 2
𝑚 (𝑡) = 𝑃
2 2 𝑚
Where 𝑃𝑚 is the modulating power. 
Input noise power is given by 2
𝑓𝑐 +𝑤

̅̅̅̅̅̅̅̅ 
𝑃𝑛𝑖 = 𝑛 2
𝑖 (𝑡) = 2 ∫ 𝑑𝑓 = 2𝑊
2
𝑓𝑐 −𝑤
−𝑓𝑐 − 𝑤 −𝑓𝑐 + 𝑤 0 𝑓𝑐 − 𝑤 𝑓𝑐 + 𝑤
Input signal to noise ratio is given by
 𝐴𝑐 2 𝑃𝑚
𝑃𝑠𝑖 2 𝐴𝑐 2 𝑃𝑚
(𝑆𝑁𝑅)𝑖 = = =
𝑃𝑛𝑖 2𝑊 4𝑊

The received signal is demodulated by multiplying r(t) by a locally generated coherent sinusoid
carrier. Then passing the product signal through a low pass filter having a bandwidth W.
𝑣(𝑡) = 𝑦(𝑡) cos(2𝜋𝑓𝑐 𝑡)

= [𝐴𝑐 𝑚(𝑡) cos(2𝜋𝑓𝑐 𝑡) + 𝑛𝑐 (𝑡) cos(2𝜋𝑓𝑐 𝑡) − 𝑛𝑠 (𝑡) sin(2𝜋𝑓𝑐 𝑡) ]cos(2𝜋𝑓𝑐 𝑡)

= 𝐴𝑐 𝑚(𝑡) cos(2𝜋𝑓𝑐 𝑡) cos(2𝜋𝑓𝑐 𝑡) + 𝑛𝑐 (𝑡) cos(2𝜋𝑓𝑐 𝑡) cos(2𝜋𝑓𝑐 𝑡) − 𝑛𝑠 (𝑡) sin(2𝜋𝑓𝑐 𝑡) cos(2𝜋𝑓𝑐 𝑡)
𝐴𝑐 𝑚(𝑡) 𝑛𝑐 (𝑡) 𝑛𝑠 (𝑡)
= [1 + cos(4𝜋𝑓𝑐 𝑡)] + [1 + cos(4𝜋𝑓𝑐 𝑡)] − sin(4𝜋𝑓𝑐 𝑡)
2 2 2
The low pass filter rejects the double frequency components (2𝑓𝑐 ) and passes only the low
frequency components. Therefore the output of LPF is given by
𝐴𝑐 𝑚(𝑡) 𝑛𝑐 (𝑡)
𝑣0 (𝑡) = +
2 2
𝑣0 (𝑡) = 𝑠0 (𝑡) + 𝑛0 (𝑡)
Therefore the output signal power is given by
𝐺𝑛c (𝑓)
̅̅̅̅̅̅̅̅
2
𝐴𝑐 2 ̅̅̅̅̅̅̅̅
2
𝐴𝑐 2
𝑃𝑠0 = 𝑠0 (𝑡) = 𝑚 (𝑡) = 𝑃
4 4 𝑚 
Where 𝑃𝑚 is the modulating power.
And output noise power is given by
𝑤
1 1 2𝑊 𝑓
𝑃𝑛0 = ̅̅̅̅̅̅̅̅
𝑛0 (𝑡)2 = ̅̅̅̅̅̅̅̅̅̅
[𝑛𝑐 (𝑡)]2 = ∗ ∫  𝑑𝑓 = −𝑊 𝑊
4 4 4
−𝑤
Therefore output SNR is given by
𝐴𝑐 2 𝑃𝑚
𝑃𝑠 4 𝐴𝑐 2 𝑃𝑚 𝑃𝑠
(𝑆𝑁𝑅)0 = 0 = 2𝑊 = = 𝑖
𝑃𝑛0 2𝑊 𝑊
4
And Noise Figure
𝐴𝑐 2 𝑃𝑚
(𝑆𝑁𝑅)0 2𝑊
𝐹𝑖𝑔𝑢𝑟𝑒 𝑜𝑓 𝑀𝑒𝑟𝑖𝑡 = = =2
(𝑆𝑁𝑅)𝑖 𝐴𝑐 2 𝑃𝑚
4𝑊
Noise in SSB system:
Let the transmitted SSB signal is
̂ (𝑡) sin(2𝜋𝑓𝑐 𝑡)
𝑠(𝑡) = 𝑚(𝑡) cos(2𝜋𝑓𝑐 𝑡) + 𝑚
The received signal at the input of the receiver is added to the white noise and filtered in a BPF.
A filtered noise can be expressed in terms of its in-phase and quadrature components as
𝑛(𝑡) = 𝑛𝑐 (𝑡) cos(2𝜋𝑓𝑐 𝑡) − 𝑛𝑠 (𝑡) sin(2𝜋𝑓𝑐 𝑡)
Let us consider SSB receiver as shown in the figure with input 𝑠(𝑡) + 𝑛(𝑡)
𝑦(𝑡) = 𝑠𝑖 (𝑡) + 𝑛𝑖 (𝑡)
Input
𝑠(𝑡) + 𝑛(𝑡) Product
BPF LPF 𝑣0 (𝑡) = 𝑠0 (𝑡) + 𝑛0 (𝑡)
Modulator

cos(2𝜋𝑓𝑐 𝑡)
After BPF the noise is bandpass noise. Then the output of BPF is sum of modulated signal and
bandpass noise with nc(t) as in-phase component and ns(t) as quadrature phase component.
Therefore the input of detector is
𝑦(𝑡) = 𝑠𝑖 (𝑡) + 𝑛𝑖 (𝑡) = 𝑚(𝑡) cos(2𝜋𝑓𝑐 𝑡) + 𝑚
̂ (𝑡) sin(2𝜋𝑓𝑐 𝑡) + 𝑛𝑐 (𝑡) cos(2𝜋𝑓𝑐 𝑡) − 𝑛𝑠 (𝑡) sin(2𝜋𝑓𝑐 𝑡)
Input signal power is given by
̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
𝑃𝑠𝑖 = [𝑚(𝑡) cos(2𝜋𝑓𝑐 𝑡) + 𝑚 ̂ (𝑡) sin(2𝜋𝑓𝑐 𝑡)]2 𝐺𝑛i (𝑓)
̅̅̅̅̅̅̅̅̅
𝑚2 (𝑡) ̅̅̅̅̅̅̅̅̅
𝑚2 (𝑡)
= + = ̅̅̅̅̅̅̅̅̅
𝑚2 (𝑡) = 𝑃𝑚
2 2 
Where 𝑃𝑚 is the modulating power. 2
Input noise power is given by
𝑓𝑐 +𝑤

𝑃𝑛𝑖 = ̅̅̅̅̅̅̅̅
𝑛𝑖 (𝑡)2 = 2 ∫ 𝑑𝑓 = 𝑊
2 −𝑓𝑐 − 𝑤 −𝑓𝑐 0 𝑓𝑐 𝑓𝑐 + 𝑤
𝑓𝑐
Input signal to noise ratio is given by
𝑃𝑠𝑖 𝑃𝑚
(𝑆𝑁𝑅)𝑖 = =
𝑃𝑛𝑖 𝑊
The received signal is demodulated by multiplying r(t) by a locally generated coherent sinusoid
carrier. Then passing the product signal through a low pass filter having a bandwidth W.
𝑣(𝑡) = 𝑦(𝑡) cos(2𝜋𝑓𝑐 𝑡)

̂ (𝑡) sin(2𝜋𝑓𝑐 𝑡) + 𝑛𝑐 (𝑡) cos(2𝜋𝑓𝑐 𝑡) − 𝑛𝑠 (𝑡) sin(2𝜋𝑓𝑐 𝑡) ]cos(2𝜋𝑓𝑐 𝑡)

= [𝑚(𝑡) cos(2𝜋𝑓𝑐 𝑡) + 𝑚
𝑚(𝑡) ̂ (𝑡)
𝑚 𝑛𝑐 (𝑡) 𝑛𝑠 (𝑡)
= [1 + cos(4𝜋𝑓𝑐 𝑡)] + sin(4𝜋𝑓𝑐 𝑡) + [1 + cos(4𝜋𝑓𝑐 𝑡)] − sin(4𝜋𝑓𝑐 𝑡)
2 2 2 2
The low pass filter rejects the double frequency components (2𝑓𝑐 ) and passes only the low
frequency components. Therefore the output of LPF is given by
𝑚(𝑡) 𝑛𝑐 (𝑡)
𝑣0 (𝑡) = +
2 2
𝑣0 (𝑡) = 𝑠0 (𝑡) + 𝑛0 (𝑡)
Therefore the output signal power is given by
̅̅̅̅̅̅̅̅
𝑚2 (𝑡) 𝑃𝑚
𝑃𝑠0 = 𝑠̅̅̅̅̅̅̅̅
2
0 (𝑡) = =
4 4
Where 𝑃𝑚 is the modulating power.
And output noise power is given by
𝑤 𝐺𝑛c (𝑓)
̅̅̅̅̅̅̅̅ 1 ̅̅̅̅̅̅̅̅̅̅2 1  𝑊
𝑃𝑛0 = 𝑛 2
0 (𝑡) = [𝑛𝑐 (𝑡)] = ∗ ∫ 𝑑𝑓 =
4 4 2 4 /2
−𝑤
Therefore output SNR is given by
𝑃𝑚
𝑃𝑠 𝑃𝑚 𝑃𝑠
(𝑆𝑁𝑅)0 = 0 = 4𝑊 = = 𝑖
𝑃𝑛0 𝑊 𝑊 𝑓
4 −𝑊 𝑊
And Noise Figure
𝑃𝑚
(𝑆𝑁𝑅)0 𝑊
𝐹𝑖𝑔𝑢𝑟𝑒 𝑜𝑓 𝑀𝑒𝑟𝑖𝑡 = = 𝑃𝑚 = 1
(𝑆𝑁𝑅)𝑖
𝑊
Noise in FM system:
Frequency modulation can be considered as a special case of phase modulation and hence we can
demodulate FM input with a PM demodulator followed by a differentiator as shown in the figure
below.
𝑦(𝑡) = 𝑠𝑖 (𝑡) + 𝑛𝑖 (𝑡) 𝑣(𝑡)
Input
𝑠(𝑡) + 𝑛(𝑡) PM 𝑑 𝑣0 (𝑡) = 𝑠0 (𝑡) + 𝑛0 (𝑡)
BPF LPF
Detector 𝑑𝑡

Let the input or received signal component at the input of detector be

𝑡
𝑠𝑖 (𝑡) = 𝐴𝑐 cos[2𝜋𝑓𝑐 𝑡 + 2𝜋𝑘𝑓 ∫−∞ 𝑚(𝑡) 𝑑𝑡]

𝑠𝑖 (𝑡) = 𝐴𝑐 cos[2𝜋𝑓𝑐 𝑡 + ∅(𝑡)]

𝑡
Where ∅(𝑡) = 2𝜋𝑘𝑓 ∫−∞ 𝑚(𝑡) 𝑑𝑡
At the input of the detector the is bandpass and given as
𝑛𝑖 (𝑡) = 𝑛𝑐 (𝑡) cos(2𝜋𝑓𝑐 𝑡) − 𝑛𝑠 (𝑡) sin(2𝜋𝑓𝑐 𝑡)
This noise is represented as a phasor like
𝑛𝑖 (𝑡) = 𝐸𝑛 cos[2𝜋𝑓𝑐 𝑡 + 𝜃(𝑡)]
Therefore the sum of signal and input noise at input of the detector is
𝑦(𝑡) = 𝑠𝑖 (𝑡) + 𝑛𝑖 (𝑡)
= 𝐴𝑐 cos[2𝜋𝑓𝑐 𝑡 + ∅(𝑡) + 𝐸𝑛 (t)cos[2𝜋𝑓𝑐 𝑡 + 𝜃(𝑡)]
To calculate the SNR at output let us add the signal and
input nose using phasor diagram at the input of the receiver.
𝐸𝑛 (𝑡)
𝑦(𝑡) = 𝐴𝑐 cos[2𝜋𝑓𝑐 𝑡 + ∅(𝑡) + ∆∅(𝑡)] 𝑅(𝑡)

Where ∆∅(𝑡) 𝜃(𝑡) − ∅(𝑡)

𝐸𝑛 (𝑡) 𝐴𝑐
∆∅(𝑡) ≈ sin[ 𝜃(𝑡) − ∅(𝑡)] ∅(𝑡) 𝜔𝑐 𝑡 + 𝜃(𝑡)
𝐴𝑐
𝑅𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒
𝜔𝑐 𝑡 + ∅(𝑡)
Therefore the detector output is
𝑣(𝑡) = ∅(𝑡) + ∆∅(𝑡)
Where
𝑡 𝐸𝑛 (𝑡)
∅(𝑡) = 2𝜋𝑘𝑓 ∫−∞ 𝑚(𝑡) 𝑑𝑡 And ∆∅(𝑡) ≈ sin[ 𝜃(𝑡) − ∅(𝑡)]
𝐴𝑐

𝐸𝑛 (𝑡) 𝐸𝑛 (𝑡) 𝐸𝑛 (𝑡)

∆∅(𝑡) ≈ sin[ 𝜃(𝑡) − 𝐾] = sin[ 𝜃(𝑡)]cos[K] − cos[ 𝜃(𝑡)]sin[𝐾]
𝐴𝑐 𝐴𝑐 𝐴𝑐
𝑛𝑐 (𝑡) 𝑛𝑠 (𝑡)
∆∅(𝑡) = cos[K] − sin[K]
𝐴𝑐 𝐴𝑐

Where
𝑛𝑐 (𝑡) = 𝐸𝑛 (𝑡)sin[ 𝜃(𝑡)] And 𝑛𝑠 (𝑡) = 𝐸𝑛 (𝑡)cos[ 𝜃(𝑡)]

Therefore power spectral density of ∆∅(𝑡) is

cos 2(K) sin2 (K)
𝐺∆∅ (𝑓) = 𝐺𝑛𝑐 (𝑓) + 𝐺𝑛𝑠 (𝑓)
𝐴2𝑐 𝐴2𝑐

𝐺𝑛𝑐 (𝑓)
= [cos2 (K) + sin2 (K)] ∵ 𝐺𝑛𝑐 (𝑓) = 𝐺𝑛𝑠 (𝑓)
𝐴2𝑐
𝐺𝑛 (𝑓)
= 𝑐2 ∵ cos 2 (K) + sin2 (K) = 1
𝐴𝑐
For white channel noise the 𝐺𝑛𝑐 (𝑓) =  𝐺 (𝑓)
 𝑛c
∴ 𝐺∆∅ (𝑓) =
𝐴2𝑐 
The output noise power spectral density 𝐺𝑛0 (𝑓) corresponding to
the input power spectral density of differentiator whose Transfer
function is 𝐻𝑑 (𝑓) = 𝑗𝜔 will be
𝑓
 −𝑊 𝑊
𝐺𝑛0 (𝑓) = 𝐺∆∅ (𝑓)|𝐻𝑑 (𝑓)|2 = 2 𝜔2
𝐴𝑐
Output noise power 𝑃𝑛0
𝑊
 8𝜋2 w3
𝑃𝑛0 = 2 ∫ 2 (2𝜋𝑓)2 𝑑𝑓 =
𝐴𝑐 0
3𝐴2𝑐
The output signal component is the output of differentiator to which the input is
𝑡
∅(𝑡) = 2𝜋𝑘𝑓 ∫−∞ 𝑚(𝑡) 𝑑𝑡 then the output signal component is
∅(𝑡) = 2𝜋𝑘𝑓 𝑘𝑑 𝑚(𝑡)
Assume 𝑘𝑑 = 1/2𝜋 then
∅(𝑡) = 𝑘𝑓 𝑚(𝑡)
Therefore output signal power 𝑃𝑠0 is
̅̅̅̅̅̅̅̅̅̅̅̅̅2
𝑃𝑠0 = [𝑘𝑓 𝑚(𝑡)] = 𝑘2𝑓 ̅̅̅̅̅̅̅̅
𝑚2 (𝑡)
 2 ̅̅̅̅̅̅̅̅
2 (𝑡)
𝑃𝑠0 𝑘𝑓 𝑚
(𝑆𝑁𝑅)0 = =
𝑃𝑛 0 8𝜋2 w3
3𝐴2𝑐

̅̅̅̅̅̅̅̅ 3𝐴2𝑐
= 𝑘2𝑓 𝑚 2 (𝑡)
8𝜋2 w3
̅̅̅̅̅̅̅̅
𝑘𝑓2 𝑚 2 (𝑡) 2
𝐴𝑐 /2 𝑘𝑓 𝑃𝑠𝑖
2
=3 = 3 ( ) 𝑃𝑚 ( )
(2𝜋𝑤)2 w 2𝜋𝑤 w

̅̅̅̅̅̅̅̅
Where 𝑃𝑚 = 𝑚 2 (𝑡) is the message signal power.

Noise in PCM System:

The output X (t) in a PCM system can be written as
𝑦(𝑡) = 𝑠0 (𝑡) + 𝑛𝑞 (𝑡) + 𝑛𝑐 (𝑡)
where 𝑠0 (𝑡) is the signal component in the output, 𝑛𝑞 (𝑡) and 𝑛𝑐 (𝑡) are two noise components.
The first noise component 𝑛𝑞 (𝑡) is due to quantization process and the noise component 𝑛𝑐 (𝑡) is
due to the channel noise. The overall signal-to-noise ratio at the output is
𝑃𝑠0 ̅̅̅̅̅̅̅̅̅̅
[𝑠0 (𝑡)]2
(𝑆𝑁𝑅)0 = =
̅̅̅̅̅̅̅̅̅̅̅2 ̅̅̅̅̅̅̅̅̅̅2
𝑃𝑛𝑞 + 𝑃𝑛𝑐 [𝑛
𝑞 (𝑡)] + [𝑛𝑐 (𝑡)]

Since these noise components are due to the errors created by the quantization and channel noise
these can be replaced by the mean square values of quantization error and channel error
respectively.
𝑃𝑠0 ̅̅̅̅̅̅̅̅̅
[𝑠(𝑡)]2
(𝑆𝑁𝑅)0 = =
𝑃𝑛𝑞 + 𝑃𝑛𝑐 ̅̅̅̅̅̅̅̅̅̅
2
[𝑒𝑞 (𝑡)] + ̅̅̅̅̅̅̅̅̅̅
[𝑒𝑐 (𝑡)]2
Quantization Noise Power:
It has been observed that the quantized signal and the original signal from which it was derived
differ one from other in a random fashion. This difference or error may be viewed as a noise due
to the quantization process and hence is called as quantization error. Let us try to estimate the
quantization error. Assume the peak to peak to range of analog message signal is divided into M
number of voltage intervals each of size  volts and their centers are quantization levels.
Therefore the quantization error or quantization noise is given by
𝑛𝑞 (𝑡) = 𝑒𝑞 (𝑡) = 𝑠(𝑡) − 𝑠𝑞 (𝑡)
That is after dropping arguments
𝑛𝑞 = 𝑒𝑞 = 𝑠 − 𝑠𝑞
∆ ∆
This quantization error (noise) varies randomly between − 2 and + 2 . Let its probability density
function be 𝑓𝑁 (𝑛) then the quantization error (noise) power is given by its mean square value as
 𝑉𝐻

𝑒̅̅̅̅
2 2
𝑞 = ∫ 𝑒𝑞 𝑓𝑁 (𝑛)𝑑𝑛
𝑉𝐿
But the analog signal amplitude rang 𝑉𝐻 − 𝑉𝐿 is divided into M number of levels the equation can
written as
∆
+
𝑀 2

𝑒̅̅̅̅
2 2
𝑞 = ∑ ∫ 𝑒𝑞 𝑓𝑘 (𝑒𝑞 )𝑑𝑒𝑞
𝑘=1 ∆
−2
∆ ∆ ∆
+ + +
2 2 2

= ∫ 𝑒𝑞 2 𝑓1 (𝑒𝑞 )𝑑𝑒𝑞 + ∫ 𝑒𝑞 2 𝑓2 (𝑒𝑞 )𝑑𝑒𝑞 +. . . . . ∫ 𝑒𝑞 2 𝑓𝑀 (𝑒𝑞 )𝑑𝑒𝑞

∆ ∆ ∆
−2 −2 −2
∆ ∆
+ +
2 2

= [𝑓1 (𝑒𝑞 ) + 𝑓2 (𝑒𝑞 ) . . . . 𝑓𝑀 (𝑒𝑞 )] ∫ 𝑒𝑞 2 𝑑𝑒𝑞 = [𝑓1 (𝑒𝑞 ) + 𝑓2 (𝑒𝑞 ) . . . . 𝑓𝑀 (𝑒𝑞 )] ∫ 𝑒𝑞 2 𝑑𝑒𝑞
∆ ∆
−2 −2
∆
+
𝑒𝑞 3 2 ∆3
= [𝑓1 (𝑒𝑞 ) + 𝑓2 (𝑒𝑞 ) . . . . 𝑓𝑀 (𝑒𝑞 )] [ ] = [𝑓1 (𝑒𝑞 ) + 𝑓2 (𝑒𝑞 ) . . . . 𝑓𝑀 (𝑒𝑞 )]
3 −∆ 12
2
∆2
= [𝑓1 (𝑒𝑞 )∆ + 𝑓2 (𝑒𝑞 ) ∆. . . . 𝑓𝑀 (𝑒𝑞 )∆]
12

Since [𝑓1 (𝑒𝑞 )∆ + 𝑓2 (𝑒𝑞 ) ∆. . . . 𝑓𝑀 (𝑒𝑞 )∆] is the total probability and equal to unity. Therefore
above equation becomes
∆2
𝑒̅̅̅̅
𝑞
2 =
12
Signal power:
Signal power is given by
𝑀∆
𝑉𝐻 +
2

𝑠̅̅̅̅̅̅̅
2 (𝑡) = ∫ 𝑠 2 𝑓 (𝑠)𝑑𝑠 = ∫ 𝑠 2 𝑓 (𝑠)𝑑𝑠
𝑆 𝑆
𝑉𝐿 𝑀∆
−2

Since x(t) is assumed to be equiprobable its probability density function 𝑓𝑋 (𝑥) can be evaluated
As shown below
𝑓𝑠 (𝑠)

𝑥
𝑀∆ 𝑀∆
− +
2 2

The total probability which is equal to the area under the distribution of 𝑓𝑠 (𝑠) must be unity.
Therefore
𝑇𝑜𝑡𝑎𝑙 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑡𝑦 = 𝐴 ∗ 𝑓𝑋 (𝑥) = 𝐴𝑀∆ = 1
1
∴ 𝑓𝑠 (𝑠) =
𝑀∆
𝑀∆
+
̅̅̅̅̅̅̅ 1 𝑠3 2 (𝑀∆)2
𝑠 2 (𝑡) = [ ] =
𝑀∆ 3 −𝑀∆ 12
2
Therefore Signal to quantization noise ratio is given by
∆2
̅̅̅̅̅̅̅̅̅
[𝑠(𝑡)]2 12
(𝑆𝑁𝑅)𝑞 = = = 𝑀2 = 22𝑁
̅̅̅̅̅̅̅̅̅̅2 (𝑀∆)2
[𝑒𝑞 (𝑡)] 12

[(𝑆𝑁𝑅)𝑞 ]𝑑𝐵 = 10 𝑙𝑜𝑔10 𝑀 = 20 𝑙𝑜𝑔10 2𝑁 = 20 ∗ 𝑁 𝑙𝑜𝑔10 2 = 6𝑁 𝑑𝐵

2

Channel noise:
In order to calculate the effects of bit errors induced by channel noise, let us consider a PCM
system using N-bit code words (M = 2N) . Let us further assume that a code word used to identify
a quantization level in the order of numerical significance of the word, that is, we assign
00…....00 to the most negative level, 00…….01 the next level, and 111…….11 to the most
positive level. An error that occurs in the least significant bit of the code words corresponds to an
error in the quantized value of the sampled signal by amount. An error in the next significant bit
causes an error of 2∆ and an error in the ith bit position causes an error of 2𝑖−1 ∆. Then the mean
square error due to channel noise is given by
𝑁 2𝑁
1 2 ∆2 2 − 1
[𝑒𝑐 (𝑡]2 = ∑ [ 2𝑖−1 ∆] =
̅̅̅̅̅̅̅̅̅
𝑁 𝑁 3
𝑖=1
For 𝑁 ≥ 2.
2𝑁 2𝑁
̅̅̅̅̅̅̅̅̅ ∆2 2 𝑃𝑒 ∆2 2
[𝑒 𝑐 (𝑡]2 ≈ = ∵ Bit error probability 𝑃𝑒 = 1/𝑁
3𝑁 3
Therefore the (SNR)0 is
̅̅̅̅̅̅̅̅̅
[𝑠(𝑡)] 2

̅̅̅̅̅̅̅̅̅
[𝑠(𝑡)]2 ̅̅̅̅̅̅̅̅̅̅̅
[𝑒𝑞 (𝑡)]
2
22𝑁
(𝑆𝑁𝑅)0 = = =
̅̅̅̅̅̅̅̅̅̅2 ̅̅̅̅̅̅̅̅̅̅2 ̅̅̅̅̅̅̅̅̅̅̅
[𝑒𝑐 (𝑡)]2 𝑃𝑒 ∆2 22𝑁 12
[𝑒𝑞 (𝑡)] + [𝑒𝑐 (𝑡)] 1+ ̅̅̅̅̅̅̅̅̅̅̅
2
1+ 3 ∆2
[𝑒𝑞 (𝑡)]

22𝑁
∴ (𝑆𝑁𝑅)0 =
1 + 4𝑃𝑒 22𝑁
For small 𝑃𝑒
(𝑆𝑁𝑅)0 = 22𝑁
For large 𝑃𝑒
1
(SNR)0 =
4Pe
But the Pe decreases with increase in (SNR)i
For large (SNR)i
(𝑆𝑁𝑅)0 = 22𝑁
For small (SNR)i
1
(SNR)0 =
4Pe

From the above relationship between (SNR)i and (SNR)0 it is clear that (SNR)0 increases
with (SNR)i for the small values of (SNR)i up to a specific value of (SNR)i and after that
(SNR)0 becomes constant that means independent of (SNR)i. This is clear from the plots shows
in figure. This effect is called Threshold effect.

Noise in DM System:
In DM system message is transmitted in the form of single bit (two level) coded pulses. Hence
after the quantization process the error 𝑒𝑞 (𝑡) will be ranging from -∆ to ∆. Therefore the mean
square value of this quantization error 𝑒𝑞 (𝑡) is given by
∆
̅̅̅̅̅̅̅̅̅̅2
[𝑒𝑞 (𝑡)] = ∫ 𝑒𝑞 2 𝑓𝑒 (𝑒𝑞 )𝑑𝑒𝑞
−∆

Where 𝑓𝑒 (𝑒𝑞 ) is probability density function of 𝑒𝑞 (𝑡) in DM system and can be found as below

𝑓𝑒 (𝑒𝑞 )

𝑒𝑞
−∆ +∆
Therefore the mean square quantization error is given by
∆
̅̅̅̅̅̅̅̅̅̅2
[𝑒𝑞 (𝑡)] = ∫ 𝑒𝑞 2 𝑓𝑒 (𝑒𝑞 )𝑑𝑒𝑞
−∆
∆ ∆
1 1 𝑒𝑞 3
= 2 ∫ 𝑒𝑞 2 𝑑𝑒𝑞 = [ ]
2∆ ∆ 3 0
0

̅̅̅̅̅̅̅̅̅̅2 ∆2
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 [𝑒𝑞 (𝑡)] =
3
This mean square error is distributed over the frequency range of sampling rate of 𝑓𝑠 𝐻𝑧. Then the
quantization noise power spectral density 𝐺𝑛𝑞 (𝑓) is given by
∆2 1 ∆2
Gnq (f) = ∗ =
3 2fs 6fs
And output quantization noise power is
fx
∆2 ∆2 fx
Pnq = ∫ df =
6fs 3 fs
−fx

Signal Power
Let the signal be
𝑠(𝑡) = 𝐴 cos(2𝜋fx t)
Therefore the signal power is
𝐴2
̅̅̅̅̅̅̅̅̅2 =
Ps = [𝑠(𝑡)]
2
But in order to avoid slope overload error the amplitude of the signal should be
∆fs
𝐴=
2𝜋fx
After substituting the value of 𝐴 in Ps

̅̅̅̅̅̅̅̅̅ ∆2 fs 2
Ps = [𝑠(𝑡)] = ( 2 ) ( )
2
8𝜋 fx
Therefore output signal to quantization noise ratio
∆2 f 2
̅̅̅̅̅̅̅̅̅
[𝑠(𝑡)]2 Ps (8𝜋2 ) (fs ) 3 fs 3
x
(𝑆𝑁𝑅)𝑞 = = = = ( 2) ( )
̅̅̅̅̅̅̅̅̅̅2 Pnq ∆2 fx 8𝜋 fx
[𝑒𝑞 (𝑡)] 3 f s
Comparison between PCM and DM
Parameter PCM DM
No of bits per sample 4 or 8 or 16 bits used per sample Only one bit per sample
Step size Depends on the number of bits Fixed can’t be varied
Band width High Low
Generation Complex Simple
SNR Good Poor
Quantization noise or Depends on number of levels Slope over load & granular
distortion used noise occurs
Sampling rate 8Khz 64-128Khz
Bit rate 7-8,thus high bit rate PCM is 1, so it is suitable for low
superior bit rate
Application Telephony Audio and speech
processing