Answer:
BW = 100 rad/s
wlow = 452.49 rad/s
whigh = 552.49 rad/s
V(jwlow) =1414.21 < 45°V
V(jwhigh) =1414.21 <-45°V
Explanation:
To calculate bandwidth we have formula
BW = 1/RC
BW = 1/ 1000x10x10^¯6
BW = 100 rad/s
We will first calculate resonant frequency and quality factor for half power frequencies.
For resonant frequency
wo = 1/(SQRT LC)
wo = 1/SQRT 400×10¯³ × 10×10^¯6
wo = 500 rad/s
For Quality
Q = wo / BW
Q = 500/100
Q = 5
wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]
wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]
wlow = 452.49 rad/s
whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]
whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]
whigh = 552.49 rad/s
We will start with admittance at lower half power frequency
Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)
Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)
Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³
Y(jwlow) = (1-j).10¯³ S
Voltage across the network is calculated by ohm's law
V(jwlow) = I/Y(jwlow)
V(jwlow) = 2/(1-j).10¯³
V(jwlow) = 1414.2 < 45°V
Now we will calculate the admittance at higher half power frequency
Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)
Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)
Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³
Y(jwhigh) = (1+j).10¯³ S
Voltage across network will be calculated by ohm's law
V(jwhigh) = I/Y(jwhigh)
V(jwhigh) = 2/(1+j).10¯³
V(jwhigh) = 1414.2 < - 45°V
