Formula Sheet: Electronics | GATE ECE (Electronics) Mock Test Series 2025 - Electronics and Communication Engineering (ECE) PDF Download
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Page 1
Amplitude Modulation :
DSB-SC :
u (t) =
m(t) cos 2p
t
Power P =
Conventioanal AM :
u (t) =
[1 + m(t)] Cos 2p
t . as long as |m(t)| = 1 demodulation is simple .
Practically m(t) = a m
(t) .
Modulation index a =
( )
( )
, m
(t) =
( )
| ( )|
Power =
+
SSB-AM :
? Square law Detector SNR =
( )
Square law modulator
?
= 2a
/ a
? amplitude Sensitivity
Envelope Detector R
C (i/p) < < 1 /
R
C (o/P) >> 1/
R
C << 1/?
=
Frequency & Phase Modulation : Angle Modulation :-
u (t) =
Cos (2p
t + Ø (t) )
Ø (t)
( ) ?
2p
m(t)
. dt ?
phase & frequency deviation constant
? max phase deviation ?Ø =
max | m(t) |
? max requency deviation ? =
max |m(t) |
Bandwidth :
Effective Bandwidth
= 2 (ß + 1)
? 98% power
Noise in Analog Modulation :-
? (SNR)
=
=
=
R = m(t) cos 2p
?
=
/ 2
Page 2
Amplitude Modulation :
DSB-SC :
u (t) =
m(t) cos 2p
t
Power P =
Conventioanal AM :
u (t) =
[1 + m(t)] Cos 2p
t . as long as |m(t)| = 1 demodulation is simple .
Practically m(t) = a m
(t) .
Modulation index a =
( )
( )
, m
(t) =
( )
| ( )|
Power =
+
SSB-AM :
? Square law Detector SNR =
( )
Square law modulator
?
= 2a
/ a
? amplitude Sensitivity
Envelope Detector R
C (i/p) < < 1 /
R
C (o/P) >> 1/
R
C << 1/?
=
Frequency & Phase Modulation : Angle Modulation :-
u (t) =
Cos (2p
t + Ø (t) )
Ø (t)
( ) ?
2p
m(t)
. dt ?
phase & frequency deviation constant
? max phase deviation ?Ø =
max | m(t) |
? max requency deviation ? =
max |m(t) |
Bandwidth :
Effective Bandwidth
= 2 (ß + 1)
? 98% power
Noise in Analog Modulation :-
? (SNR)
=
=
=
R = m(t) cos 2p
?
=
/ 2
? (SNR)
=
/
/
=
=
=
=
= (SNR)
? (SNR)
=
/
/
=
=
=
= (SNR)
.
=
.
= ?
? =
Noise in Angle Modulation :-
=
PCM :-
? Min. no of samples required for reconstruction = 2? =
; ? = Bandwidth of msg signal .
? Total bits required = v
bps . v ? bits / sample
? Bandwidth = R
/2 = v
/ 2 = v . ?
? SNR = 1.76 + 6.02 v
? As Number of bits increased SNR increased by 6 dB/bit . Band width also increases.
Delta Modulation :-
? By increasing step size slope over load distortion eliminated [ Signal raised sharply ]
? By Reducing step size Grannualar distortion eliminated . [ Signal varies slowly ]
Digital Communication
Matched filter:
? impulse response a(t) =
( T – t) . P(t) ? i/p
? Matched filter o/p will be max at multiples of ‘T’ . So, sampling @ multiples of ‘T’ will give max SNR
(2
nd
point )
? matched filter is always causal a(t) = 0 for t < 0
? Spectrum of o/p signal of matched filter with the matched signal as i/p ie, except for a delay factor ;
proportional to energy spectral density of i/p.
Ø
( ) =
(f) Ø(f) = Ø(f) Ø*(f) e
Ø
( ) = |Ø( )|
e
Page 3
Amplitude Modulation :
DSB-SC :
u (t) =
m(t) cos 2p
t
Power P =
Conventioanal AM :
u (t) =
[1 + m(t)] Cos 2p
t . as long as |m(t)| = 1 demodulation is simple .
Practically m(t) = a m
(t) .
Modulation index a =
( )
( )
, m
(t) =
( )
| ( )|
Power =
+
SSB-AM :
? Square law Detector SNR =
( )
Square law modulator
?
= 2a
/ a
? amplitude Sensitivity
Envelope Detector R
C (i/p) < < 1 /
R
C (o/P) >> 1/
R
C << 1/?
=
Frequency & Phase Modulation : Angle Modulation :-
u (t) =
Cos (2p
t + Ø (t) )
Ø (t)
( ) ?
2p
m(t)
. dt ?
phase & frequency deviation constant
? max phase deviation ?Ø =
max | m(t) |
? max requency deviation ? =
max |m(t) |
Bandwidth :
Effective Bandwidth
= 2 (ß + 1)
? 98% power
Noise in Analog Modulation :-
? (SNR)
=
=
=
R = m(t) cos 2p
?
=
/ 2
? (SNR)
=
/
/
=
=
=
=
= (SNR)
? (SNR)
=
/
/
=
=
=
= (SNR)
.
=
.
= ?
? =
Noise in Angle Modulation :-
=
PCM :-
? Min. no of samples required for reconstruction = 2? =
; ? = Bandwidth of msg signal .
? Total bits required = v
bps . v ? bits / sample
? Bandwidth = R
/2 = v
/ 2 = v . ?
? SNR = 1.76 + 6.02 v
? As Number of bits increased SNR increased by 6 dB/bit . Band width also increases.
Delta Modulation :-
? By increasing step size slope over load distortion eliminated [ Signal raised sharply ]
? By Reducing step size Grannualar distortion eliminated . [ Signal varies slowly ]
Digital Communication
Matched filter:
? impulse response a(t) =
( T – t) . P(t) ? i/p
? Matched filter o/p will be max at multiples of ‘T’ . So, sampling @ multiples of ‘T’ will give max SNR
(2
nd
point )
? matched filter is always causal a(t) = 0 for t < 0
? Spectrum of o/p signal of matched filter with the matched signal as i/p ie, except for a delay factor ;
proportional to energy spectral density of i/p.
Ø
( ) =
(f) Ø(f) = Ø(f) Ø*(f) e
Ø
( ) = |Ø( )|
e
? o/p signal of matched filter is proportional to shifted version of auto correlation fine of i/p signal
Ø
(t) = R
Ø
(t – T)
At t = T Ø
(T) = R
Ø
(0) ? which proves 2
nd
point
Cauchy-Schwartz in equality :-
|g
(t) g
(t) dt|
= g
(t)
dt |g
(t)|
dt
If g
(t) = c g
(t) then equality holds otherwise ‘<’ holds
Raised Cosine pulses :
P(t) =
(
)
(
)
.
(
)
P(f) =
| | =
cos
| |
=| | =
| |
? Bamdwidth of Raised cosine filter
=
? Bit rate
=
a ? roll o actor
? signal time period
? For Binary PSK
= Q
= Q
=
erfc
.
? 4 PSK
= 2Q
1
FSK:-
For BPSK
= Q
= Q
=
erfc
? All signals have same energy (Const energy modulation )
? Energy & min distance both can be kept constant while increasing no. of points . But Bandwidth
Compramised.
? PPM is called as Dual of FSK .
? For DPSK
=
e
/
Page 4
Amplitude Modulation :
DSB-SC :
u (t) =
m(t) cos 2p
t
Power P =
Conventioanal AM :
u (t) =
[1 + m(t)] Cos 2p
t . as long as |m(t)| = 1 demodulation is simple .
Practically m(t) = a m
(t) .
Modulation index a =
( )
( )
, m
(t) =
( )
| ( )|
Power =
+
SSB-AM :
? Square law Detector SNR =
( )
Square law modulator
?
= 2a
/ a
? amplitude Sensitivity
Envelope Detector R
C (i/p) < < 1 /
R
C (o/P) >> 1/
R
C << 1/?
=
Frequency & Phase Modulation : Angle Modulation :-
u (t) =
Cos (2p
t + Ø (t) )
Ø (t)
( ) ?
2p
m(t)
. dt ?
phase & frequency deviation constant
? max phase deviation ?Ø =
max | m(t) |
? max requency deviation ? =
max |m(t) |
Bandwidth :
Effective Bandwidth
= 2 (ß + 1)
? 98% power
Noise in Analog Modulation :-
? (SNR)
=
=
=
R = m(t) cos 2p
?
=
/ 2
? (SNR)
=
/
/
=
=
=
=
= (SNR)
? (SNR)
=
/
/
=
=
=
= (SNR)
.
=
.
= ?
? =
Noise in Angle Modulation :-
=
PCM :-
? Min. no of samples required for reconstruction = 2? =
; ? = Bandwidth of msg signal .
? Total bits required = v
bps . v ? bits / sample
? Bandwidth = R
/2 = v
/ 2 = v . ?
? SNR = 1.76 + 6.02 v
? As Number of bits increased SNR increased by 6 dB/bit . Band width also increases.
Delta Modulation :-
? By increasing step size slope over load distortion eliminated [ Signal raised sharply ]
? By Reducing step size Grannualar distortion eliminated . [ Signal varies slowly ]
Digital Communication
Matched filter:
? impulse response a(t) =
( T – t) . P(t) ? i/p
? Matched filter o/p will be max at multiples of ‘T’ . So, sampling @ multiples of ‘T’ will give max SNR
(2
nd
point )
? matched filter is always causal a(t) = 0 for t < 0
? Spectrum of o/p signal of matched filter with the matched signal as i/p ie, except for a delay factor ;
proportional to energy spectral density of i/p.
Ø
( ) =
(f) Ø(f) = Ø(f) Ø*(f) e
Ø
( ) = |Ø( )|
e
? o/p signal of matched filter is proportional to shifted version of auto correlation fine of i/p signal
Ø
(t) = R
Ø
(t – T)
At t = T Ø
(T) = R
Ø
(0) ? which proves 2
nd
point
Cauchy-Schwartz in equality :-
|g
(t) g
(t) dt|
= g
(t)
dt |g
(t)|
dt
If g
(t) = c g
(t) then equality holds otherwise ‘<’ holds
Raised Cosine pulses :
P(t) =
(
)
(
)
.
(
)
P(f) =
| | =
cos
| |
=| | =
| |
? Bamdwidth of Raised cosine filter
=
? Bit rate
=
a ? roll o actor
? signal time period
? For Binary PSK
= Q
= Q
=
erfc
.
? 4 PSK
= 2Q
1
FSK:-
For BPSK
= Q
= Q
=
erfc
? All signals have same energy (Const energy modulation )
? Energy & min distance both can be kept constant while increasing no. of points . But Bandwidth
Compramised.
? PPM is called as Dual of FSK .
? For DPSK
=
e
/
? Orthogonal signals require factor of ‘2’ more energy to achieve same
as anti podal signals
? Orthogonal signals are 3 dB poorer than antipodal signals. The 3dB difference is due to distance b/w 2
points.
? For non coherent FSK
=
e
/
? FPSK & 4 QAM both have comparable performance .
? 32 QAM has 7 dB advantage over 32 PSK.
? Bandwidth of Mary PSK =
=
; S =
? Bandwidth of Mary FSK =
=
; S =
? Bandwidth efficiency S =
.
.
? Symbol time
=
log
? Band rate =
Page 5
Amplitude Modulation :
DSB-SC :
u (t) =
m(t) cos 2p
t
Power P =
Conventioanal AM :
u (t) =
[1 + m(t)] Cos 2p
t . as long as |m(t)| = 1 demodulation is simple .
Practically m(t) = a m
(t) .
Modulation index a =
( )
( )
, m
(t) =
( )
| ( )|
Power =
+
SSB-AM :
? Square law Detector SNR =
( )
Square law modulator
?
= 2a
/ a
? amplitude Sensitivity
Envelope Detector R
C (i/p) < < 1 /
R
C (o/P) >> 1/
R
C << 1/?
=
Frequency & Phase Modulation : Angle Modulation :-
u (t) =
Cos (2p
t + Ø (t) )
Ø (t)
( ) ?
2p
m(t)
. dt ?
phase & frequency deviation constant
? max phase deviation ?Ø =
max | m(t) |
? max requency deviation ? =
max |m(t) |
Bandwidth :
Effective Bandwidth
= 2 (ß + 1)
? 98% power
Noise in Analog Modulation :-
? (SNR)
=
=
=
R = m(t) cos 2p
?
=
/ 2
? (SNR)
=
/
/
=
=
=
=
= (SNR)
? (SNR)
=
/
/
=
=
=
= (SNR)
.
=
.
= ?
? =
Noise in Angle Modulation :-
=
PCM :-
? Min. no of samples required for reconstruction = 2? =
; ? = Bandwidth of msg signal .
? Total bits required = v
bps . v ? bits / sample
? Bandwidth = R
/2 = v
/ 2 = v . ?
? SNR = 1.76 + 6.02 v
? As Number of bits increased SNR increased by 6 dB/bit . Band width also increases.
Delta Modulation :-
? By increasing step size slope over load distortion eliminated [ Signal raised sharply ]
? By Reducing step size Grannualar distortion eliminated . [ Signal varies slowly ]
Digital Communication
Matched filter:
? impulse response a(t) =
( T – t) . P(t) ? i/p
? Matched filter o/p will be max at multiples of ‘T’ . So, sampling @ multiples of ‘T’ will give max SNR
(2
nd
point )
? matched filter is always causal a(t) = 0 for t < 0
? Spectrum of o/p signal of matched filter with the matched signal as i/p ie, except for a delay factor ;
proportional to energy spectral density of i/p.
Ø
( ) =
(f) Ø(f) = Ø(f) Ø*(f) e
Ø
( ) = |Ø( )|
e
? o/p signal of matched filter is proportional to shifted version of auto correlation fine of i/p signal
Ø
(t) = R
Ø
(t – T)
At t = T Ø
(T) = R
Ø
(0) ? which proves 2
nd
point
Cauchy-Schwartz in equality :-
|g
(t) g
(t) dt|
= g
(t)
dt |g
(t)|
dt
If g
(t) = c g
(t) then equality holds otherwise ‘<’ holds
Raised Cosine pulses :
P(t) =
(
)
(
)
.
(
)
P(f) =
| | =
cos
| |
=| | =
| |
? Bamdwidth of Raised cosine filter
=
? Bit rate
=
a ? roll o actor
? signal time period
? For Binary PSK
= Q
= Q
=
erfc
.
? 4 PSK
= 2Q
1
FSK:-
For BPSK
= Q
= Q
=
erfc
? All signals have same energy (Const energy modulation )
? Energy & min distance both can be kept constant while increasing no. of points . But Bandwidth
Compramised.
? PPM is called as Dual of FSK .
? For DPSK
=
e
/
? Orthogonal signals require factor of ‘2’ more energy to achieve same
as anti podal signals
? Orthogonal signals are 3 dB poorer than antipodal signals. The 3dB difference is due to distance b/w 2
points.
? For non coherent FSK
=
e
/
? FPSK & 4 QAM both have comparable performance .
? 32 QAM has 7 dB advantage over 32 PSK.
? Bandwidth of Mary PSK =
=
; S =
? Bandwidth of Mary FSK =
=
; S =
? Bandwidth efficiency S =
.
.
? Symbol time
=
log
? Band rate =
? Energy of a signal |x(t)|
dt = | [ ]|
? Power of a signal P = lim
?
|x(t)|
dt = lim
?
|x[n]|
? x
(t) ?
; x
(t) ?
x
(t) + x
(t) ?
+
iff x
(t) & x
(t) orthogonal
? Shifting & Time scaling won’t effect power . Frequency content doesn’t effect power.
? if power = 8 ? neither energy nor power signal
Power = 0 ? Energy signal
Power = K ? power signal
? Energy of power signal = 8 ; Power of energy signal = 0
? Generally Periodic & random signals ? Power signals
Aperiodic & deterministic ? Energy signals
Precedence rule for scaling & Shifting :
x(at + b) ? (1) shift x(t) by ‘b’ ? x(t + b)
(2) Scale x(t + b) by ‘a’ ? x(at + b)
x( a ( t + b/a)) ? (1) scale x(t) by a ? x(at)
(2) shift x(at) by b/a ? x (a (t+b/a)).
? x(at +b) = y(t) ? x(t) = y
? Step response s(t) = h(t) * u(t) = h(t)dt
S’ (t) = h(t)
S[n] = [ ]
h[n] = s[n] – s[n-1]
? e
u(t) * e
u(t) =
[ e
- e
] u(t) .
?
Rect (t / 2
) *
Rect(t / 2
) = 2
min (
,
) trapezoid (
,
)
? Rect (t / 2T) * Rect (t / 2T) = 2T tri(t / T)
Hilbert Transform Pairs :
e
/
dx
= s 2p ; x
e
/
dx = s
2p s > 0
Laplace Transform :-
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FAQs on Formula Sheet: Electronics - GATE ECE (Electronics) Mock Test Series 2025 - Electronics and Communication Engineering (ECE)
| 1. What are some common electronic components used in Electronics and Communication Engineering? |  |
Ans. Some common electronic components used in Electronics and Communication Engineering are resistors, capacitors, inductors, transistors, diodes, and integrated circuits.
| 2. How does a transistor work in electronic circuits? | |
Ans. A transistor is a semiconductor device that can amplify or switch electronic signals. It works by controlling the flow of current between its terminals based on the voltage applied to its input terminal.
| 3. What is the purpose of using capacitors in electronic circuits? | |
Ans. Capacitors are used in electronic circuits to store and release electrical energy, filter out noise and ripple in power supplies, and block DC signals while allowing AC signals to pass through.
| 4. How do integrated circuits (ICs) differ from discrete electronic components? | |
Ans. Integrated circuits (ICs) are miniature electronic circuits fabricated on a small piece of semiconductor material, while discrete electronic components are individual components like resistors and capacitors that are not integrated into a single package.
| 5. How do communication systems use modulation and demodulation techniques? | |
Ans. Modulation is the process of encoding information onto a carrier signal, while demodulation is the process of extracting the original information from the modulated signal. Communication systems use modulation and demodulation techniques to transmit and receive information efficiently over long distances.
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Oct 23, 2024 Last updated |
Document Description: Formula Sheet: Electronics for Electronics and Communication Engineering (ECE) 2024 is part of GATE ECE (Electronics) Mock Test Series 2025 preparation.
The notes and questions for Formula Sheet: Electronics have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus. Information about Formula Sheet: Electronics covers topics
like and Formula Sheet: Electronics Example, for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Formula Sheet: Electronics.
Introduction of Formula Sheet: Electronics in English is available as part of our GATE ECE (Electronics) Mock Test Series 2025
for Electronics and Communication Engineering (ECE) & Formula Sheet: Electronics in Hindi for GATE ECE (Electronics) Mock Test Series 2025 course.
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Formula Sheet: Electronics Free PDF Download
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