E72(a) Review, Solutions

 

Problem 1:

  1. R3, R5 and R6 are in series (Rx=R3+R5+R6),
    and this is in parallel with R4 (Ry=Rx||R4),
    and this is in series with R2 (Rz=R2+Ry),
    and this is in parallel with R1,
      RAB=Rz||R1=636 Ohm
            
  2. R1 is in series with R2 and in parallel with R4 (Rx=(R1+R2)||R4,
    and this is in series with R5 and R6 and in parallel with R3
       RCD=(Rx+R5+R6)||R3=727 Ohm
             
  3. The series combination of R3, R5 and R6 is in parallel with R4, (Rx=(R3+R4+R5)||R4),
    and this is in parallel with the series combination of R1 and R2,
       RBC=(R1+R2)||Rx = 545 Ohm
             

Problem 2:

  1. There are only independent sources so the Thevenin Resistance is 3k Ohm (=1k+1k+2k||2k).  The Thevenin Voltage is 4.5 volts (=2.5+2).
  2. Thevenin Resistance is 1k Ohm (=2k||2k), and the Thevenin voltage is 3.5 volts (=2.5+1).

Problem 3:  This one is hard, but relates directly to some transistor concepts.

  1. Apply a test voltage across nodes A and B, and find resulting current
  2. Apply a test voltage across nodes C and D, and find resulting current.

Problem 4: 

  1. As seen from Rx, the Thevenin equivalent is Vt=9.5 V(=5+4.5), Rt=100 Ohm.  Therefore, for max power, Rx=100 Ohm.
  2. Power dissipated in Rx is equal to the power in the other three resistors.  Power=4.75V*47.5mA=.23 W.

Problem 5: 

For both parts of this problem t=RC and v(t)=v(∞)+(v(0)-v(∞))e-t/t

  1. (a) v(0)=0, v(∞)=1.     b) v(0)=1, v(∞)=0.
  2. (a) v(0)=1, v(∞)=-1.    b) v(0)=-2, v(∞)=0.

Problem 6: 

a) 


  1. Exact in blue, asymptotic in red
  2. From Bode plot (asymptotic) magnitude is unity (0 dB) and phase is 0 (-6o on exact plot), so output is equal to input (10sin(100t).
  3. From Bode plot, magnitude is 0.1 (-20 dB) and phase is -90, so output is sin(10000t-90°)=-cos(10000t).
  4. This is a lowpass filter

b) 


  1. Exact in blue, asymptotic in red
  2. From Bode plot, magnitude is 0.1(-20 dB) and phase is 90, so output is sin(100t+90°)=cos(100t).
  3. From Bode plot, magnitude is unity and phase is 0, so output is 10sin(10000t).
  4. This is a highpass filter

Problem 7: 

For both parts of this problem t=RC and v(t)=v(∞)+(v(0)-v(∞))e-t/t

  1. When the switch is opened
  2. When the switch is closed

Problem 8: 

For both parts of this problem t=RC and v(t)=v(∞)+(v(0)-v(∞))e-t/t

  1. When the switch is opened
  2. When the switch is closed

Problem 9: 


Problem 10: 

Recall that for a generic first order system with a step input that:

so

 


Problem 11: 

  1. At low frequencies, inductor is short, output is zero.  At high frequencies, capacitor is short, output is zero.  Output must be non-zero at intermediate frequencies, therefore this is a bandpass filter.
  2. At s=0 and s=∞, the magnitude of the transfer function is zero.
  3. It decreases.

Problem 12: 

  1. At low frequencies, capacitor is open circuit, output=input.  At high frequencies, capacitor is short circuit, output=0.  Therefore this is a lowpass filter.
  2. At s=0, Xfer function=1.  At s=∞, the Xfer function=0. 
  3. It increases.

Problem 13: 

  1. The expression below shows Vo=0 if r=0
  2. see (i).
  3. if r<<R, then

Problem 14: 

  1. 2.5V+2V=4.5V
  2. 2.5V+1V=3.5V

Problem 15: 

We take the positive root since resistance must be positive.