356791
An oscillator is producing FM waves of \(A M\) wire is \(24 \mathrm{mV}\) and the minimum peak to peak voltage is \(8 \mathrm{mV}\). The modulation factor is
1 \(20 \%\)
2 \(10 \%\)
3 \(50 \%\)
4 \(25 \%\)
Explanation:
Here, \({V_{\max }} = \frac{{24}}{2} = 12mV\) and \({V_{\min }} = \frac{8}{2} = 2mV\) Now, \(m = \frac{{{V_{\max {\rm{ }}}} - {V_{\min {\rm{ }}}}}}{{{V_{\max {\rm{ }}}} + {V_{\min {\rm{ }}}}}}\) \(=\dfrac{12-4}{12+4}=\dfrac{8}{16}=\dfrac{1}{2}=0.5=50 \%\) \(h_{R}\) are doubled then s bemomes \(\sqrt{2}\) times.
PHXII15:COMMUNICATION SYSTEMS
356792
The antenna current of an AM transmitter is \(8 \mathrm{~A}\) when only the carrier is sent, but it increases to \(8.93 \mathrm{~A}\) when the carrier is modulated by a single wave. The percentage modulation is
356793
If \(f_{c}\) and \(f_{m}\) are the frequencies of carrier wave and signal, then the bandwidth is
1 \(f_{m}\)
2 \(2 f_{m}\)
3 \(f_{c}\)
4 \(2 f_{c}\)
Explanation:
Conceptual Question
PHXII15:COMMUNICATION SYSTEMS
356794
A modulating signal is a square wave, as shown in the figure.
If the carrier wave is given as \(C(t) = 2\sin \,(8\,\pi t){\text{volts}},\) the modulation index is
1 \(\dfrac{1}{3}\)
2 1
3 \(\dfrac{1}{2}\)
4 \(\dfrac{1}{4}\)
Explanation:
From the given diagram maximum of \(m(t)=A_{m}=1 {~V}\) Also, maximum of \(c(t)=A_{c}=2 {~V}\) \(\therefore\) Modulation index \(\mu=\dfrac{A_{m}}{A_{c}}=\dfrac{1}{2}\) So option (3) is correct.
356791
An oscillator is producing FM waves of \(A M\) wire is \(24 \mathrm{mV}\) and the minimum peak to peak voltage is \(8 \mathrm{mV}\). The modulation factor is
1 \(20 \%\)
2 \(10 \%\)
3 \(50 \%\)
4 \(25 \%\)
Explanation:
Here, \({V_{\max }} = \frac{{24}}{2} = 12mV\) and \({V_{\min }} = \frac{8}{2} = 2mV\) Now, \(m = \frac{{{V_{\max {\rm{ }}}} - {V_{\min {\rm{ }}}}}}{{{V_{\max {\rm{ }}}} + {V_{\min {\rm{ }}}}}}\) \(=\dfrac{12-4}{12+4}=\dfrac{8}{16}=\dfrac{1}{2}=0.5=50 \%\) \(h_{R}\) are doubled then s bemomes \(\sqrt{2}\) times.
PHXII15:COMMUNICATION SYSTEMS
356792
The antenna current of an AM transmitter is \(8 \mathrm{~A}\) when only the carrier is sent, but it increases to \(8.93 \mathrm{~A}\) when the carrier is modulated by a single wave. The percentage modulation is
356793
If \(f_{c}\) and \(f_{m}\) are the frequencies of carrier wave and signal, then the bandwidth is
1 \(f_{m}\)
2 \(2 f_{m}\)
3 \(f_{c}\)
4 \(2 f_{c}\)
Explanation:
Conceptual Question
PHXII15:COMMUNICATION SYSTEMS
356794
A modulating signal is a square wave, as shown in the figure.
If the carrier wave is given as \(C(t) = 2\sin \,(8\,\pi t){\text{volts}},\) the modulation index is
1 \(\dfrac{1}{3}\)
2 1
3 \(\dfrac{1}{2}\)
4 \(\dfrac{1}{4}\)
Explanation:
From the given diagram maximum of \(m(t)=A_{m}=1 {~V}\) Also, maximum of \(c(t)=A_{c}=2 {~V}\) \(\therefore\) Modulation index \(\mu=\dfrac{A_{m}}{A_{c}}=\dfrac{1}{2}\) So option (3) is correct.
356791
An oscillator is producing FM waves of \(A M\) wire is \(24 \mathrm{mV}\) and the minimum peak to peak voltage is \(8 \mathrm{mV}\). The modulation factor is
1 \(20 \%\)
2 \(10 \%\)
3 \(50 \%\)
4 \(25 \%\)
Explanation:
Here, \({V_{\max }} = \frac{{24}}{2} = 12mV\) and \({V_{\min }} = \frac{8}{2} = 2mV\) Now, \(m = \frac{{{V_{\max {\rm{ }}}} - {V_{\min {\rm{ }}}}}}{{{V_{\max {\rm{ }}}} + {V_{\min {\rm{ }}}}}}\) \(=\dfrac{12-4}{12+4}=\dfrac{8}{16}=\dfrac{1}{2}=0.5=50 \%\) \(h_{R}\) are doubled then s bemomes \(\sqrt{2}\) times.
PHXII15:COMMUNICATION SYSTEMS
356792
The antenna current of an AM transmitter is \(8 \mathrm{~A}\) when only the carrier is sent, but it increases to \(8.93 \mathrm{~A}\) when the carrier is modulated by a single wave. The percentage modulation is
356793
If \(f_{c}\) and \(f_{m}\) are the frequencies of carrier wave and signal, then the bandwidth is
1 \(f_{m}\)
2 \(2 f_{m}\)
3 \(f_{c}\)
4 \(2 f_{c}\)
Explanation:
Conceptual Question
PHXII15:COMMUNICATION SYSTEMS
356794
A modulating signal is a square wave, as shown in the figure.
If the carrier wave is given as \(C(t) = 2\sin \,(8\,\pi t){\text{volts}},\) the modulation index is
1 \(\dfrac{1}{3}\)
2 1
3 \(\dfrac{1}{2}\)
4 \(\dfrac{1}{4}\)
Explanation:
From the given diagram maximum of \(m(t)=A_{m}=1 {~V}\) Also, maximum of \(c(t)=A_{c}=2 {~V}\) \(\therefore\) Modulation index \(\mu=\dfrac{A_{m}}{A_{c}}=\dfrac{1}{2}\) So option (3) is correct.
356791
An oscillator is producing FM waves of \(A M\) wire is \(24 \mathrm{mV}\) and the minimum peak to peak voltage is \(8 \mathrm{mV}\). The modulation factor is
1 \(20 \%\)
2 \(10 \%\)
3 \(50 \%\)
4 \(25 \%\)
Explanation:
Here, \({V_{\max }} = \frac{{24}}{2} = 12mV\) and \({V_{\min }} = \frac{8}{2} = 2mV\) Now, \(m = \frac{{{V_{\max {\rm{ }}}} - {V_{\min {\rm{ }}}}}}{{{V_{\max {\rm{ }}}} + {V_{\min {\rm{ }}}}}}\) \(=\dfrac{12-4}{12+4}=\dfrac{8}{16}=\dfrac{1}{2}=0.5=50 \%\) \(h_{R}\) are doubled then s bemomes \(\sqrt{2}\) times.
PHXII15:COMMUNICATION SYSTEMS
356792
The antenna current of an AM transmitter is \(8 \mathrm{~A}\) when only the carrier is sent, but it increases to \(8.93 \mathrm{~A}\) when the carrier is modulated by a single wave. The percentage modulation is
356793
If \(f_{c}\) and \(f_{m}\) are the frequencies of carrier wave and signal, then the bandwidth is
1 \(f_{m}\)
2 \(2 f_{m}\)
3 \(f_{c}\)
4 \(2 f_{c}\)
Explanation:
Conceptual Question
PHXII15:COMMUNICATION SYSTEMS
356794
A modulating signal is a square wave, as shown in the figure.
If the carrier wave is given as \(C(t) = 2\sin \,(8\,\pi t){\text{volts}},\) the modulation index is
1 \(\dfrac{1}{3}\)
2 1
3 \(\dfrac{1}{2}\)
4 \(\dfrac{1}{4}\)
Explanation:
From the given diagram maximum of \(m(t)=A_{m}=1 {~V}\) Also, maximum of \(c(t)=A_{c}=2 {~V}\) \(\therefore\) Modulation index \(\mu=\dfrac{A_{m}}{A_{c}}=\dfrac{1}{2}\) So option (3) is correct.
356791
An oscillator is producing FM waves of \(A M\) wire is \(24 \mathrm{mV}\) and the minimum peak to peak voltage is \(8 \mathrm{mV}\). The modulation factor is
1 \(20 \%\)
2 \(10 \%\)
3 \(50 \%\)
4 \(25 \%\)
Explanation:
Here, \({V_{\max }} = \frac{{24}}{2} = 12mV\) and \({V_{\min }} = \frac{8}{2} = 2mV\) Now, \(m = \frac{{{V_{\max {\rm{ }}}} - {V_{\min {\rm{ }}}}}}{{{V_{\max {\rm{ }}}} + {V_{\min {\rm{ }}}}}}\) \(=\dfrac{12-4}{12+4}=\dfrac{8}{16}=\dfrac{1}{2}=0.5=50 \%\) \(h_{R}\) are doubled then s bemomes \(\sqrt{2}\) times.
PHXII15:COMMUNICATION SYSTEMS
356792
The antenna current of an AM transmitter is \(8 \mathrm{~A}\) when only the carrier is sent, but it increases to \(8.93 \mathrm{~A}\) when the carrier is modulated by a single wave. The percentage modulation is
356793
If \(f_{c}\) and \(f_{m}\) are the frequencies of carrier wave and signal, then the bandwidth is
1 \(f_{m}\)
2 \(2 f_{m}\)
3 \(f_{c}\)
4 \(2 f_{c}\)
Explanation:
Conceptual Question
PHXII15:COMMUNICATION SYSTEMS
356794
A modulating signal is a square wave, as shown in the figure.
If the carrier wave is given as \(C(t) = 2\sin \,(8\,\pi t){\text{volts}},\) the modulation index is
1 \(\dfrac{1}{3}\)
2 1
3 \(\dfrac{1}{2}\)
4 \(\dfrac{1}{4}\)
Explanation:
From the given diagram maximum of \(m(t)=A_{m}=1 {~V}\) Also, maximum of \(c(t)=A_{c}=2 {~V}\) \(\therefore\) Modulation index \(\mu=\dfrac{A_{m}}{A_{c}}=\dfrac{1}{2}\) So option (3) is correct.
