ocs.google.com/forms/d/e/1FAIPQLSFM signal is better than AM signal because *O stereo techniques may applyO more immune to noiseO All of the aboveO Less adjacent channel interferenceThe signal can be reconstructedO At Nyquist rateO Above Nyquist rateO At & above the Nyquist rateO Below Nyquist ratePAM signal can be extracted by using.O LPFO HPFO PLLO PWM

Advantage of FM compare to AM These are modulation technique. as per our guidelines we are…

Answered: google.com/torms/d/e/1FAIPQL FM signal…

google.com/torms/d/e/1FAIPQL FM signal is better than AM signal because * stereo techniques may apply O more immune to noise O All of the above O Less adjacent channel interference

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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ı just need fınal aswer In the circuit in the figure, Vcc = 18 V, Rs1 = 0.9 kΩ, Rs2 = 0.5 kΩ, R1 = 75 kΩ, R2 = 15 kΩ, RC = 2.2 kΩ, RE = 1 kΩ and β =120. It is also known that Cwi = 6 pF, Cwo = 10 pF, Cbc = Cu = 7 pF, Cbe = Cπ = 58 pF and Cce = 11 pF. Accordingly, considering the frequency response of the circuit, what would be the high cutoff frequency (fH)? NOTE-1: The output impedance of the transistor is r0 = 20 kΩ and will be considered in the calculations. NOTE-2: The high frequency dependence of hfe, that is β, will be taken into account. a. 189,95 kHz b. 617,33 kHz c. 569,84 kHz d. 474,87 kHz e. 332,41 kHz f. 47,49 kHz g. 142,46 kHz h. 712,30 kHz
1. The CT signal x(t)=3cos (600πt) + 6cos2(mπt) is critically sampled (Nyquist rate) by fs=4000 Hz. Find:i. The value of m.ii. The frequency components of the signal.iii. The frequency resolution (∆f) of the output spectrum if the sampled signal is processed by 64 point-DFT.iv. Define frequency resolution.
In the circuit in the figure, Vcc= 18 V, Rs1 = 0.9 kohm, Rohm = 0.5 kQ, R1 = 75 kohm, R2 = 15 kohm, RC = 2.2 kohm, RE = 1 kohm and B = 120. It is also known that Cwi = 6 pF, Cwo = 10 pF, Cbc = Cu = 7 pF, Cbe = C = 44 pF and Cce = 11 PF. Accordingly, considering the frequency response of the circuit, what would the high cutoff frequency (fH) be? NOTE-1: The output impedance of the transistor is ro = 20 k and will be taken into account in the calculations. NOTE-2: high frequency dependence of hfe, B. will be taken into account.
1. The CT signal x(t)=3cos (600πt) + 6cos2 (mπt) is critically sampled (Nyquist rate)by fs=4000 Hz. Find:i. The value of m.ii. The frequency components of the signal.iii. The frequency resolution (Δf) of the output spectrum if the sampled signal isprocessed by 64 point-DFT.iv. Define frequency resolution.
In the circuit in the figure, Vcc= 18 V, Rs1 = 0.9 k2, Rs2 = 0.5 kQ, R1 = 75 kQ, R2 = 15 kQ, RC = 2.2 kQ, RE = 1 k and B = 120. It is also known that Cwi = 6 pF, Cwo = 10 pF, Cbc = Cu = 7 pF, Cbe = C = 44 pF and Cce = 11 PF. Accordingly, considering the frequency response of the circuit, what would the high cutoff frequency (fH) be? NOTE-1: The output impedance of the transistor is ro = 20 k and will be taken into account in the calculations. NOTE-2: high frequency dependence of hfe, B. will be taken into account.
In the circuit in the figure, Vcc = 18 V, Rs1 = 0.9 kΩ, Rs2 = 0.5 kΩ, R1 = 75 kΩ, R2 = 15 kΩ, RC = 2.2 kΩ, RE = 1 kΩ and β =120. It is also known that Cwi = 6 pF, Cwo = 10 pF, Cbc = Cu = 7 pF, Cbe = Cπ = 51 pF and Cce = 11 pF. Accordingly, considering the frequency response of the circuit, what would be the high cutoff frequency (fH)? NOTE-1: The output impedance of the transistor is r0 = 20 kΩ and will be taken into account in the calculations. NOTE-2: The high frequency dependence of hfe, β, will not be taken into account.
In the circuit in the figure, Vcc = 18 V, Rs1 = 0.9 kΩ, Rs2 = 0.5 kΩ, R1 = 75 kΩ, R2 = 15 kΩ, RC = 2.2 kΩ, RE = 1 kΩ and β =120. It is also known that Cwi = 6 pF, Cwo = 10 pF, Cbc = Cu = 7 pF, Cbe = Cπ = 48 pF and Cce = 11 pF. Accordingly, considering the frequency response of the circuit, what would be the high cutoff frequency (fH)? NOTE-1: The output impedance of the transistor is r0 = 20 kΩ and will be taken into account in the calculations. NOTE-2: The high frequency dependence of hfe, β, will be taken into account. a. 719,95 kHz b. 143,99 kHz c. 48,00 kHz d. 191,99 kHz e. 575,96 kHz f. 479,97 kHz g. 335,98 kHz h. 623,96 kHz
TRUE or FALSE 1.) In time domain represenatation of a signal, a signal with an expression of 220 cos (377t -180) is also equal to -220 cos (377t) 2.) In time domain represenatation of a signal, a signal with an expression of 220 sin (377t +90) is also equal to -220 cos (377t)
(can you solve it fast?) Which ones are correct about Nyquist's Theorem in sampling? I. Relates the conditions on the time and frequency planes ii. Helps quantize iii. They determine the bandwidth requirement iv. Returns the spectrum of the sign
In the figure, the input signal of the system is x (t) and the output signal is y (t). In this system, there is a sub-system with frequency response H (f). The frequency response H (f) of this subsystem is given at the top of the figure and defined as c = 800, k = 2. In addition, the frequency of the cosine signal shown in the figure is given as fc = 4000.x(t)= 4cos(8000πt).cos(1200πt)Find the output signal Y (t) and calculate the value at t = 1/4 , Y = 1/8 =?Given that the input signal of this system x (t) is defined as
3. The __________ is, in essence, a switch which closes for a brief period oftime and then open, closing and opening many times a second. 4. The ___________ is a way to tell how many frequency components are neededto make up a signal.
The signal s(n)s(n) is recorded with frequency sampling Fs=500HzFs=500Hz. What's the duration of the following part s1(n)s1(n) of the signal s(n)s(n) where s1=s(601:4:5000)