1- Determine the implementation complexity of the TS switch shown in the figure, where the number of TDM input lines is N = 120. Assume each input line contains a single T1 signal (24 channels). Furthermore, assume a one stage matrix is used for the space stage. Time stage Space stage 1 TSM 2 TSM N TSM N BT NxN Control store Nx=120²=14400 NBx= =(120) (24) (7)=20160 = (120) (24) (8)+(120)(24)(5)=37440 N +N Bx complexity=N+ Bt=14976 100 N 2- Determine the implementation complexity of the TS switch of exercise 1, if groups of 4 input links are combined into higher level multiplexed signals before being switched, forming 30 input links with 96 channels (time slots) per link.

Consider the line of code below. analogWrite(11, 84); The voltage at pin 11 is turns on (at 5V for Uno) and off at 500 Hz. This PWM frequency is the default value for Arduino boards.  How many microseconds is the voltage at pin 11 high?

Why is a high percentage of modulation desirable?

7+6) This QUESTION FROM DIGITAL COMMUNICATIONS II course.

19. Gate A has tPLH = tPHL = 6 ns. Gate B has tPLH = tPHL = 10 ns. Which gate can be operated at a higher frequency? Explain!

Fast please @sync1

In three-level PWM, the modulating process operates on each portion of the modulating function separately. The output peak value is still given by the depth of modulation, but the RMS value of the square wave is not Vin. Compute the THD for three-level PWM as a function of the depth of modulation (modulation index k).

Draw the DC circuit:  b)DC computations: Compute the DC collector current of each transistor. I_C 1 = ________________________ mA I_C 2= ________________________ Calculations:  c)Draw the AC circuit and compute gm, r_pi for the system. g_m 1= _______________________ mA/V r_pi 1 = _______________________ kohms g_m 2= _______________________ mA/V r_pi 2 = _______________________ kohms AC circuit:  Calculations: d)Compute the gain A_v=Vo/Vi at the operational band (for sufficiently high frequency) Gain = A_v = ______________________________ Computations:

Incremental shaft encoders have usually two pairs of photo emitters and photo detectors. Why is this? 1. Using two pairs of emitters and detectors allows turning an incremental shaftencoder into an absolute one.2. Usually the two emitter/detector pairs are out of phase by an angle of 90 degrees,so that the direction in which the the encoder disk is moving can be determined. Ifthere was only one such emitter/detector pair, the direction could not be determined.3. The additional emitter/detector pair is only introduced, in order to increase theredundancy of the shaft encoder. Hence, if one of them fails, the shaft encoder wouldcontinue operating in the same way as beforehand.4. The additional emitter/detector pair allows measuring additional information aboutthe shaft encoder motion. Whilst shaft encoders with a single emitter/detector paircan strictly only measure the absolute angle of the shaft encoder, adding another pairallows measuring differential speed and acceleration.

to continue the solution..  answer part ( vi and v ) thank you ..