Why can baseband pulses not be sent directly on radio or satellite channels?

These channels pass only a band around a centre frequency, not frequencies near DC

Baseband pulses are too weak in amplitude

Digital pulses cannot be amplified

Which statement about the envelope is correct?

FSK and PSK have constant envelopes, so resist amplitude noise better than ASK

2064 Shrawan (CS I) · 3+5 marks

Need for shift keying; ASK vs FSK vs PSK

Short answer

Shift keying transmits digital data over bandpass channels, enabling small antennas, multiplexing, and noise control. ASK varies amplitude (poor noise immunity), FSK varies frequency (constant envelope, better than ASK), PSK varies phase (constant envelope, best noise performance but requires coherent detection).

Why do we use shift keying technique in communication system? Distinguish between ASK, FSK and PSK.

Why shift keying is used

Shift keying (digital carrier modulation) means switching the amplitude, frequency or phase of a high-frequency sinusoidal carrier according to the binary data. It is used because:

Bandpass channels: radio, satellite, microwave and telephone channels pass only a band of frequencies around some centre frequency; baseband pulses (with energy near DC) cannot go through them directly.
Practical antenna size: antennas must be about $\lambda/4$ long; a high carrier frequency makes them small.
Multiplexing: different users can be given different carrier frequencies (FDM).
Noise and interference control: FSK and PSK keep a constant envelope and so resist amplitude noise, and the scheme can be chosen to trade bandwidth for power.
Long-distance transmission over wireless and existing analog lines (modems).

ASK vs FSK vs PSK

Point	ASK	FSK	PSK
Parameter changed	Amplitude	Frequency	Phase
Signal for 1 / 0	$A\cos\omega_ct$ / 0	$A\cos\omega_1t$ / $A\cos\omega_2t$	$A\cos\omega_ct$ / $-A\cos\omega_ct$
Envelope	Varies (on-off)	Constant	Constant
Bandwidth (approx.)	$2R_b$	$\lvert f_1-f_2\rvert+2R_b$ (largest)	$2R_b$
Noise immunity	Poor	Better than ASK	Best
$P_e$ (coherent)	$\tfrac12\text{erfc}\sqrt{E_b/2N_0}$	$\tfrac12\text{erfc}\sqrt{E_b/2N_0}$	$\tfrac12\text{erfc}\sqrt{E_b/N_0}$
Detection	Envelope or coherent	Non-coherent or coherent	Coherent only (or DPSK)
Complexity	Simplest	Moderate	Most complex (carrier recovery)
Signal space	1-D, points 0 and $\sqrt{2E_b}$	2-D, orthogonal points	1-D, antipodal $\pm\sqrt{E_b}$
Typical use	Optical fibre, low-rate links	Low-speed modems, paging	Satellite, Wi-Fi, high-speed modems

 data :   1      0      1
 ASK  : /\/\/\ ______ /\/\/\
 FSK  : /\/\/\ /\  /\ /\/\/\   (fewer cycles for 0)
 PSK  : /\/\/\ \/\/\/ /\/\/\   (phase flips)

( $E_b$ is the average energy per bit; for ASK the "1" bit has energy $2E_b$ . On this basis coherent ASK has the same $P_e$ as coherent FSK, but its decision threshold depends on the received amplitude, so it suffers more from amplitude noise and fading. If $E_b$ is instead taken as the energy of the "1" bit, the ASK result is written $\tfrac12\text{erfc}\sqrt{E_b/4N_0}$ .)

✓ Answer checked by Claude Opus 5.5 (high effort)

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