#1. An AM transmitter has envelope extremes Vmax = 24 V and Vmin = 4 V at the detector. If carrier power is 800 W, what is the total transmitted power?
#2. An AM transmitter has envelope extremes Vmax = 22 V and Vmin = 4 V at the detector. If carrier power is 1000 W, what is the total transmitted power?
#3. An AM transmitter has envelope extremes Vmax = 20 V and Vmin = 12 V at the detector. If carrier power is 1800 W, what is the total transmitted power?
#4. An AM transmitter has envelope extremes Vmax = 24 V and Vmin = 8 V at the detector. If carrier power is 1500 W, what is the total transmitted power?
#5. An AM transmitter has envelope extremes Vmax = 24 V and Vmin = 6 V at the detector. If carrier power is 1800 W, what is the total transmitted power?
#6. An AM transmitter has envelope extremes Vmax = 26 V and Vmin = 6 V at the detector. If carrier power is 1000 W, what is the total transmitted power?
#7. An AM transmitter has envelope extremes Vmax = 26 V and Vmin = 6 V at the detector. If carrier power is 1500 W, what is the total transmitted power?
#8. An FM signal has peak frequency deviation ±75 kHz and maximum modulating frequency 3 kHz. Using Carson’s rule, what transmission bandwidth is required?
#9. An FM signal has peak frequency deviation ±45 kHz and maximum modulating frequency 3 kHz. Using Carson’s rule, what transmission bandwidth is required?
#10. An FM signal has peak frequency deviation ±50 kHz and maximum modulating frequency 12 kHz. Using Carson’s rule, what transmission bandwidth is required?
#11. An FM signal has peak frequency deviation ±60 kHz and maximum modulating frequency 10 kHz. Using Carson’s rule, what transmission bandwidth is required?
#12. An FM signal has peak frequency deviation ±50 kHz and maximum modulating frequency 10 kHz. Using Carson’s rule, what transmission bandwidth is required?
#13. An FM signal has peak frequency deviation ±75 kHz and maximum modulating frequency 3 kHz. Using Carson’s rule, what transmission bandwidth is required?
#14. A PCM-TDM system carries 32 channels. Each analog channel has highest frequency 5 kHz, sampling is performed at 1.5×Nyquist, and quantization is 8 bits/sample. If framing overhead is 4%, what is aggregate line rate?
#15. A PCM-TDM system carries 30 channels. Each analog channel has highest frequency 3.4 kHz, sampling is performed at 1.1×Nyquist, and quantization is 8 bits/sample. If framing overhead is 4%, what is aggregate line rate?
#16. A PCM-TDM system carries 24 channels. Each analog channel has highest frequency 4 kHz, sampling is performed at 1.2×Nyquist, and quantization is 8 bits/sample. If framing overhead is 5%, what is aggregate line rate?
#17. A PCM-TDM system carries 32 channels. Each analog channel has highest frequency 3.4 kHz, sampling is performed at 1.1×Nyquist, and quantization is 12 bits/sample. If framing overhead is 4%, what is aggregate line rate?
#18. A PCM-TDM system carries 24 channels. Each analog channel has highest frequency 6 kHz, sampling is performed at 1.5×Nyquist, and quantization is 12 bits/sample. If framing overhead is 3%, what is aggregate line rate?
#19. A PCM-TDM system carries 24 channels. Each analog channel has highest frequency 5 kHz, sampling is performed at 1.25×Nyquist, and quantization is 10 bits/sample. If framing overhead is 3%, what is aggregate line rate?
#20. A conventional AM transmitter operates with modulation index m = 0.3 and total radiated power 2000 W. What is sideband power conversion efficiency η = Psb/Pt?
#21. A conventional AM transmitter operates with modulation index m = 0.9 and total radiated power 2400 W. What is sideband power conversion efficiency η = Psb/Pt?
#22. A conventional AM transmitter operates with modulation index m = 0.9 and total radiated power 2400 W. What is sideband power conversion efficiency η = Psb/Pt?
#23. A conventional AM transmitter operates with modulation index m = 0.3 and total radiated power 1500 W. What is sideband power conversion efficiency η = Psb/Pt?
#24. A conventional AM transmitter operates with modulation index m = 0.5 and total radiated power 2000 W. What is sideband power conversion efficiency η = Psb/Pt?
#25. A conventional AM transmitter operates with modulation index m = 0.8 and total radiated power 1800 W. What is sideband power conversion efficiency η = Psb/Pt?
#26. For a periodic signal x(t) with period T = 4 ms, a delayed version y(t)=x(t-0.5 ms) is formed. What is the phase shift of the n=10 Fourier-series harmonic in y(t) relative to x(t)?
#27. For a periodic signal x(t) with period T = 0.5 ms, a delayed version y(t)=x(t-0.2 ms) is formed. What is the phase shift of the n=10 Fourier-series harmonic in y(t) relative to x(t)?
#28. For a periodic signal x(t) with period T = 1 ms, a delayed version y(t)=x(t-0.5 ms) is formed. What is the phase shift of the n=6 Fourier-series harmonic in y(t) relative to x(t)?
#29. For a periodic signal x(t) with period T = 8 ms, a delayed version y(t)=x(t-0.1 ms) is formed. What is the phase shift of the n=4 Fourier-series harmonic in y(t) relative to x(t)?
#30. For a periodic signal x(t) with period T = 1 ms, a delayed version y(t)=x(t-0.05 ms) is formed. What is the phase shift of the n=6 Fourier-series harmonic in y(t) relative to x(t)?
#31. For a periodic signal x(t) with period T = 4 ms, a delayed version y(t)=x(t-0.5 ms) is formed. What is the phase shift of the n=3 Fourier-series harmonic in y(t) relative to x(t)?
#32. For a periodic signal x(t) with period T = 2 ms, a delayed version y(t)=x(t-0.05 ms) is formed. What is the phase shift of the n=6 Fourier-series harmonic in y(t) relative to x(t)?
#33. For a periodic signal x(t) with period T = 8 ms, a delayed version y(t)=x(t-0.2 ms) is formed. What is the phase shift of the n=10 Fourier-series harmonic in y(t) relative to x(t)?
#34. A periodic waveform is x(t)= 2.5 + 2cos(ω0t)+4sin(ω0t)+3cos(2ω0t)+3sin(2ω0t). Using Parseval’s theorem, what is x_rms?
#35. A periodic waveform is x(t)= 2.5 + 5cos(ω0t)+2sin(ω0t)+3cos(2ω0t)+2sin(2ω0t). Using Parseval’s theorem, what is x_rms?
#36. A periodic waveform is x(t)= 2.5 + 3cos(ω0t)+1sin(ω0t)+1cos(2ω0t)+3sin(2ω0t). Using Parseval’s theorem, what is x_rms?
#37. A periodic waveform is x(t)= 2.5 + 2cos(ω0t)+4sin(ω0t)+1cos(2ω0t)+3sin(2ω0t). Using Parseval’s theorem, what is x_rms?
#38. A periodic waveform is x(t)= 2.5 + 3cos(ω0t)+3sin(ω0t)+2cos(2ω0t)+1sin(2ω0t). Using Parseval’s theorem, what is x_rms?
#39. A periodic waveform is x(t)= 2 + 3cos(ω0t)+2sin(ω0t)+3cos(2ω0t)+2sin(2ω0t). Using Parseval’s theorem, what is x_rms?
#40. A periodic waveform is x(t)= 1.5 + 3cos(ω0t)+4sin(ω0t)+2cos(2ω0t)+3sin(2ω0t). Using Parseval’s theorem, what is x_rms?
#41. A unipolar rectangular pulse train has pulse width τ=50 μs and duty cycle 20%. What harmonic index n gives the first spectral null (n = f_null/f0)?
#42. A unipolar rectangular pulse train has pulse width τ=80 μs and duty cycle 50%. What harmonic index n gives the first spectral null (n = f_null/f0)?
#43. A unipolar rectangular pulse train has pulse width τ=25 μs and duty cycle 40%. What harmonic index n gives the first spectral null (n = f_null/f0)?
#44. A unipolar rectangular pulse train has pulse width τ=40 μs and duty cycle 50%. What harmonic index n gives the first spectral null (n = f_null/f0)?
#45. A unipolar rectangular pulse train has pulse width τ=40 μs and duty cycle 50%. What harmonic index n gives the first spectral null (n = f_null/f0)?
#46. A signal x(t) has spectrum X(jω)=1 for |ω|≤400 rad/s and 0 otherwise. For y(t)=x(4t), evaluate |Y(jω)| at ω=125 rad/s.
#47. A signal x(t) has spectrum X(jω)=1 for |ω|≤300 rad/s and 0 otherwise. For y(t)=x(4t), evaluate |Y(jω)| at ω=50 rad/s.
#48. A signal x(t) has spectrum X(jω)=1 for |ω|≤400 rad/s and 0 otherwise. For y(t)=x(5t), evaluate |Y(jω)| at ω=100 rad/s.
#49. A signal x(t) has spectrum X(jω)=1 for |ω|≤500 rad/s and 0 otherwise. For y(t)=x(3t), evaluate |Y(jω)| at ω=125 rad/s.
#50. A signal x(t) has spectrum X(jω)=1 for |ω|≤200 rad/s and 0 otherwise. For y(t)=x(4t), evaluate |Y(jω)| at ω=125 rad/s.
#51. A synchronous TDM system multiplexes 48 PCM channels, each sampled at 16 ksample/s with 10 bits/sample. Each frame adds 4 framing bits, and line coding expansion factor is 2. What is final line rate?
#52. A synchronous TDM system multiplexes 30 PCM channels, each sampled at 16 ksample/s with 12 bits/sample. Each frame adds 2 framing bits, and line coding expansion factor is 2. What is final line rate?
#53. A synchronous TDM system multiplexes 24 PCM channels, each sampled at 8 ksample/s with 12 bits/sample. Each frame adds 4 framing bits, and line coding expansion factor is 1.25. What is final line rate?
#54. A synchronous TDM system multiplexes 32 PCM channels, each sampled at 10 ksample/s with 10 bits/sample. Each frame adds 6 framing bits, and line coding expansion factor is 1.25. What is final line rate?
#55. A synchronous TDM system multiplexes 48 PCM channels, each sampled at 10 ksample/s with 10 bits/sample. Each frame adds 6 framing bits, and line coding expansion factor is 1. What is final line rate?
#56. A synchronous TDM system multiplexes 30 PCM channels, each sampled at 10 ksample/s with 8 bits/sample. Each frame adds 1 framing bits, and line coding expansion factor is 2. What is final line rate?
#57. A synchronous TDM system multiplexes 48 PCM channels, each sampled at 12 ksample/s with 8 bits/sample. Each frame adds 2 framing bits, and line coding expansion factor is 1. What is final line rate?
#58. A synchronous TDM system multiplexes 30 PCM channels, each sampled at 16 ksample/s with 10 bits/sample. Each frame adds 4 framing bits, and line coding expansion factor is 1. What is final line rate?
#59. An FDM trunk carries 8 channels each of 25 kHz. Adjacent channels require 1.5 kHz guard bands, and one 12 kHz pilot tone is inserted. What is total occupied bandwidth?
#60. An FDM trunk carries 12 channels each of 25 kHz. Adjacent channels require 2.5 kHz guard bands, and one 10 kHz pilot tone is inserted. What is total occupied bandwidth?
#61. An FDM trunk carries 10 channels each of 30 kHz. Adjacent channels require 3 kHz guard bands, and one 10 kHz pilot tone is inserted. What is total occupied bandwidth?
#62. An FDM trunk carries 16 channels each of 12.5 kHz. Adjacent channels require 2.5 kHz guard bands, and one 5 kHz pilot tone is inserted. What is total occupied bandwidth?
#63. An FDM trunk carries 12 channels each of 30 kHz. Adjacent channels require 2.5 kHz guard bands, and one 8 kHz pilot tone is inserted. What is total occupied bandwidth?
#64. An FDM trunk carries 8 channels each of 25 kHz. Adjacent channels require 1 kHz guard bands, and one 10 kHz pilot tone is inserted. What is total occupied bandwidth?
#65. An FDM trunk carries 16 channels each of 12.5 kHz. Adjacent channels require 3 kHz guard bands, and one 5 kHz pilot tone is inserted. What is total occupied bandwidth?
#66. An FDM trunk carries 16 channels each of 20 kHz. Adjacent channels require 1.5 kHz guard bands, and one 10 kHz pilot tone is inserted. What is total occupied bandwidth?
#67. In a DS-CDMA cell, chip rate is 5 Mcps and user data rate is 128 kb/s. Assuming required Eb/N0=7 dB and loading factor α=0.5, estimate user capacity K≈1+Gp/(Eb/N0·α), where Gp=Rc/Rb.
#68. In a DS-CDMA cell, chip rate is 5 Mcps and user data rate is 256 kb/s. Assuming required Eb/N0=6 dB and loading factor α=0.35, estimate user capacity K≈1+Gp/(Eb/N0·α), where Gp=Rc/Rb.
#69. In a DS-CDMA cell, chip rate is 1.2288 Mcps and user data rate is 256 kb/s. Assuming required Eb/N0=5 dB and loading factor α=0.45, estimate user capacity K≈1+Gp/(Eb/N0·α), where Gp=Rc/Rb.
#70. In a DS-CDMA cell, chip rate is 5 Mcps and user data rate is 19.2 kb/s. Assuming required Eb/N0=8 dB and loading factor α=0.45, estimate user capacity K≈1+Gp/(Eb/N0·α), where Gp=Rc/Rb.
#71. In a DS-CDMA cell, chip rate is 10 Mcps and user data rate is 64 kb/s. Assuming required Eb/N0=7 dB and loading factor α=0.45, estimate user capacity K≈1+Gp/(Eb/N0·α), where Gp=Rc/Rb.
#72. In a DS-CDMA cell, chip rate is 5 Mcps and user data rate is 9.6 kb/s. Assuming required Eb/N0=4 dB and loading factor α=0.6, estimate user capacity K≈1+Gp/(Eb/N0·α), where Gp=Rc/Rb.
#73. In a DS-CDMA cell, chip rate is 5 Mcps and user data rate is 128 kb/s. Assuming required Eb/N0=8 dB and loading factor α=0.6, estimate user capacity K≈1+Gp/(Eb/N0·α), where Gp=Rc/Rb.
#74. In a DS-CDMA cell, chip rate is 5 Mcps and user data rate is 64 kb/s. Assuming required Eb/N0=6 dB and loading factor α=0.6, estimate user capacity K≈1+Gp/(Eb/N0·α), where Gp=Rc/Rb.
#75. In a DS-CDMA cell, chip rate is 3.84 Mcps and user data rate is 64 kb/s. Assuming required Eb/N0=8 dB and loading factor α=0.45, estimate user capacity K≈1+Gp/(Eb/N0·α), where Gp=Rc/Rb.
#76. A 4-QAM link carries net data rate 20 Mb/s with raised-cosine roll-off α=0.35. What minimum RF bandwidth is required (ideal coherent detection)?
#77. A 32-QAM link carries net data rate 80 Mb/s with raised-cosine roll-off α=0.25. What minimum RF bandwidth is required (ideal coherent detection)?
#78. A 64-QAM link carries net data rate 120 Mb/s with raised-cosine roll-off α=0.25. What minimum RF bandwidth is required (ideal coherent detection)?
#79. A 8-QAM link carries net data rate 40 Mb/s with raised-cosine roll-off α=0.2. What minimum RF bandwidth is required (ideal coherent detection)?
#80. A 16-QAM link carries net data rate 10 Mb/s with raised-cosine roll-off α=0.2. What minimum RF bandwidth is required (ideal coherent detection)?
#81. A 4-QAM link carries net data rate 10 Mb/s with raised-cosine roll-off α=0.5. What minimum RF bandwidth is required (ideal coherent detection)?
#82. A 16-QAM link carries net data rate 10 Mb/s with raised-cosine roll-off α=0.25. What minimum RF bandwidth is required (ideal coherent detection)?
#83. For coherent BPSK in AWGN, BER = Q(√(2Eb/N0)). If Eb/N0 = 9 dB, what is the approximate BER?
#84. For coherent BPSK in AWGN, BER = Q(√(2Eb/N0)). If Eb/N0 = 4 dB, what is the approximate BER?
#85. For coherent BPSK in AWGN, BER = Q(√(2Eb/N0)). If Eb/N0 = 9 dB, what is the approximate BER?
#86. For coherent BPSK in AWGN, BER = Q(√(2Eb/N0)). If Eb/N0 = 8 dB, what is the approximate BER?
#87. For coherent BPSK in AWGN, BER = Q(√(2Eb/N0)). If Eb/N0 = 6 dB, what is the approximate BER?
#88. For coherent BPSK in AWGN, BER = Q(√(2Eb/N0)). If Eb/N0 = 9 dB, what is the approximate BER?
#89. A digital receiver has Pt=23 dBm, Gt=5 dBi, Gr=5 dBi, path loss=100 dB, noise figure=5 dB, and bandwidth=2 MHz. Using thermal noise -174 dBm/Hz and Shannon formula, what is channel capacity?
#90. A digital receiver has Pt=30 dBm, Gt=8 dBi, Gr=2 dBi, path loss=90 dB, noise figure=5 dB, and bandwidth=5 MHz. Using thermal noise -174 dBm/Hz and Shannon formula, what is channel capacity?
#91. A digital receiver has Pt=20 dBm, Gt=10 dBi, Gr=8 dBi, path loss=95 dB, noise figure=4 dB, and bandwidth=2 MHz. Using thermal noise -174 dBm/Hz and Shannon formula, what is channel capacity?
#92. A digital receiver has Pt=30 dBm, Gt=10 dBi, Gr=2 dBi, path loss=105 dB, noise figure=6 dB, and bandwidth=5 MHz. Using thermal noise -174 dBm/Hz and Shannon formula, what is channel capacity?
#93. A digital receiver has Pt=20 dBm, Gt=2 dBi, Gr=0 dBi, path loss=100 dB, noise figure=4 dB, and bandwidth=5 MHz. Using thermal noise -174 dBm/Hz and Shannon formula, what is channel capacity?
#94. A digital receiver has Pt=23 dBm, Gt=5 dBi, Gr=0 dBi, path loss=105 dB, noise figure=4 dB, and bandwidth=1 MHz. Using thermal noise -174 dBm/Hz and Shannon formula, what is channel capacity?
#95. A receiver measures SNR = 12 dB over noise bandwidth 3 MHz while data rate is 8 Mb/s. What is Eb/N0?
#96. A receiver measures SNR = 10 dB over noise bandwidth 1.2 MHz while data rate is 6 Mb/s. What is Eb/N0?
#97. A receiver measures SNR = 8 dB over noise bandwidth 2 MHz while data rate is 6 Mb/s. What is Eb/N0?
#98. A receiver measures SNR = 12 dB over noise bandwidth 1 MHz while data rate is 6 Mb/s. What is Eb/N0?
#99. A receiver measures SNR = 8 dB over noise bandwidth 0.8 MHz while data rate is 8 Mb/s. What is Eb/N0?
#100. A receiver measures SNR = 8 dB over noise bandwidth 1 MHz while data rate is 2 Mb/s. What is Eb/N0?
