ASK Transmitter MCQ Quiz - Objective Question with Answer for ASK Transmitter - Download Free PDF
Last updated on Apr 20, 2025
Latest ASK Transmitter MCQ Objective Questions
ASK Transmitter Question 1:
In ______ a Carrier is transmitted for a 1-bit, and nothing is transmitted for 0-bit, this is analogous to flashing light communication.
Answer (Detailed Solution Below)
ASK Transmitter Question 1 Detailed Solution
ASK System:
1. For ASK Transmitter on-off keying is used.
2. In Amplitude Shift Keying (ASK) binary 1 is represented with the presence of carrier and binary 0 is represented with the absence of the carrier.
1 : s1(t) = Ac cos 2πfct
0 : s2 (t) = 0
ASK Transmitter Question 2:
Which of the following modulation schemes gives the maximum probability of error?
Answer (Detailed Solution Below)
ASK Transmitter Question 2 Detailed Solution
Analysis:
The probability of error for ASK, PSK, and FSK is given as
\(P_{e_{ASK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{4N_0}}})} \space \)
\(P_{e_{PSK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{N_0}}})} \space \)
\(P_{e_{FSK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{2N_0}}})} \space \)
Q(x) is a decreasing function therefore as x increases the value of Q(x) decreases
\( \sqrt {\dfrac {A^2_c T_b \space }{N_0}} \space > \space \sqrt {\dfrac {A^2_c T_b \space }{2N_0}} \space > \space \sqrt {\dfrac {A^2_c T_b \space }{4N_0}} \space \)
Therefore,
\( Q (\sqrt {\dfrac {A^2_c T_b \space }{4N_0}}) \space > \space Q(\sqrt {\dfrac {A^2_c T_b \space }{2N_0}}) \space > \space Q( \sqrt {\dfrac {A^2_c T_b \space }{N_0}}) \space \)
Pe ASK > PeFSK > Pe PSK
ASK modulation scheme gives Maximum probability of error.
Top ASK Transmitter MCQ Objective Questions
In ______ a Carrier is transmitted for a 1-bit, and nothing is transmitted for 0-bit, this is analogous to flashing light communication.
Answer (Detailed Solution Below)
ASK Transmitter Question 3 Detailed Solution
Download Solution PDFASK System:
1. For ASK Transmitter on-off keying is used.
2. In Amplitude Shift Keying (ASK) binary 1 is represented with the presence of carrier and binary 0 is represented with the absence of the carrier.
1 : s1(t) = Ac cos 2πfct
0 : s2 (t) = 0
Which of the following modulation schemes gives the maximum probability of error?
Answer (Detailed Solution Below)
ASK Transmitter Question 4 Detailed Solution
Download Solution PDFAnalysis:
The probability of error for ASK, PSK, and FSK is given as
\(P_{e_{ASK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{4N_0}}})} \space \)
\(P_{e_{PSK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{N_0}}})} \space \)
\(P_{e_{FSK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{2N_0}}})} \space \)
Q(x) is a decreasing function therefore as x increases the value of Q(x) decreases
\( \sqrt {\dfrac {A^2_c T_b \space }{N_0}} \space > \space \sqrt {\dfrac {A^2_c T_b \space }{2N_0}} \space > \space \sqrt {\dfrac {A^2_c T_b \space }{4N_0}} \space \)
Therefore,
\( Q (\sqrt {\dfrac {A^2_c T_b \space }{4N_0}}) \space > \space Q(\sqrt {\dfrac {A^2_c T_b \space }{2N_0}}) \space > \space Q( \sqrt {\dfrac {A^2_c T_b \space }{N_0}}) \space \)
Pe ASK > PeFSK > Pe PSK
ASK modulation scheme gives Maximum probability of error.
ASK Transmitter Question 5:
In ______ a Carrier is transmitted for a 1-bit, and nothing is transmitted for 0-bit, this is analogous to flashing light communication.
Answer (Detailed Solution Below)
ASK Transmitter Question 5 Detailed Solution
ASK System:
1. For ASK Transmitter on-off keying is used.
2. In Amplitude Shift Keying (ASK) binary 1 is represented with the presence of carrier and binary 0 is represented with the absence of the carrier.
1 : s1(t) = Ac cos 2πfct
0 : s2 (t) = 0
ASK Transmitter Question 6:
Which of the following modulation schemes gives the maximum probability of error?
Answer (Detailed Solution Below)
ASK Transmitter Question 6 Detailed Solution
Analysis:
The probability of error for ASK, PSK, and FSK is given as
\(P_{e_{ASK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{4N_0}}})} \space \)
\(P_{e_{PSK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{N_0}}})} \space \)
\(P_{e_{FSK}} = Q ( {{\sqrt {\dfrac {A^2_c T_b \space }{2N_0}}})} \space \)
Q(x) is a decreasing function therefore as x increases the value of Q(x) decreases
\( \sqrt {\dfrac {A^2_c T_b \space }{N_0}} \space > \space \sqrt {\dfrac {A^2_c T_b \space }{2N_0}} \space > \space \sqrt {\dfrac {A^2_c T_b \space }{4N_0}} \space \)
Therefore,
\( Q (\sqrt {\dfrac {A^2_c T_b \space }{4N_0}}) \space > \space Q(\sqrt {\dfrac {A^2_c T_b \space }{2N_0}}) \space > \space Q( \sqrt {\dfrac {A^2_c T_b \space }{N_0}}) \space \)
Pe ASK > PeFSK > Pe PSK
ASK modulation scheme gives Maximum probability of error.
ASK Transmitter Question 7:
In ASK system, pulse waveform are defined as;
\(\rm S_1(t)=\sqrt2 \ sin\left[\frac{\pi t}{4T}\right]mV\) 0 ≤ t ≤ 4T; bit 1
S2 (t) = 0; 4T ≤ t ≤ 8T ; bit 0
Find probability of error, if symbols are equiprobable and noise is AWGN with two sided PSD is 10-12 W/Hz. [assume Tb = 1 μsec]
Answer (Detailed Solution Below)
ASK Transmitter Question 7 Detailed Solution
Concept:
"If symbols are equiprobable ⇒ coherent detection"
In ASK, when coherent detection is done, then
Probability of error; \(\rm P_e=Q\left[√{{\frac{d^2min}{2No}}}\right]\)
where d2min = Eb (bit energy)
No = PSD
Calculation:
For bit 1
⇒ A = √2 × 10-3 volt
∴ \(\rm E_b=\frac{A^2}{2}× T_b=\frac{(√2×10^{-3})^2}{2}×10^{-6}\)
= 10-12 J
∴ Two sided PSD ⇒ \(\rm \frac{No}{2}=10^{-12}\)
No = 2 × 10-12 w/Hz
∴ \(\rm P_e=Q\left[√{\frac{E_b}{2No}}\right]\)
= Q[0.5]
Note:-
for coherent Detection for equiprobable symbol;
\(\rm P_e=Q\left[√{\frac{d^2min}{2No}}\right]\)
dmin = √Eb ---- for ASK
dmin = √2Eb ---- for FSK
dmin = 2√Eb ---- for PSK
For Non-coherent Detection
\(\rm P_e=\frac{1}{2}e^{\frac{-E_b}{No}}\) --- for PSK
\(\rm P_e=\frac{1}{2}e^{\frac{-E_b}{2No}}\) ---- for ASK and FSK