Final answer:
The bandwidth of the bandpass filter with a center frequency of 50krad/s and a quality factor of 4 is 1.99 kHz. The upper cutoff frequency is approximately 8.95 kHz, and the lower cutoff frequency is around 6.96 kHz, both expressed in kilohertz.
Step-by-step explanation:
To find the bandwidth of a bandpass filter, we use the relationship between the quality factor (Q), the resonant or center frequency (ω0), and the bandwidth (BW). The formula given is BW = ω0 / Q. Given a center frequency of 50krad/s and a quality factor of 4, we can calculate the bandwidth as 50krad/s / 4 = 12.5krad/s. To convert krad/s to kilohertz (kHz), we divide by 2π because 1 rad/s = 1/2π Hz. Therefore, the bandwidth in kHz is approximately 12.5k / 2π = 1.99 kHz.
Now, to find the upper and lower cutoff frequencies, we use the fact that the bandwidth is the difference between these two frequencies. The center frequency is the average of the upper (ωH) and lower (ωL) cutoff frequencies. Thus, ω0 = (ωH + ωL) / 2. Since we already know the bandwidth, we can solve for the cutoff frequencies: ωH = ω0 + (BW/2) and ωL = ω0 - (BW/2). Plugging in our values, we get the upper cutoff frequency as approximately (50 + 6.25)k rad/s = 56.25k rad/s or 8.95 kHz, and the lower cutoff frequency as approximately (50 - 6.25)k rad/s = 43.75k rad/s or 6.96 kHz.