Unity gain in lowpass STC circuit

[gambit1414]

One of the question asks to find the frequency at which the gain becomes 0dB for a low pass Single Time Constant circuit. In the solution manual after finding the 3dB frequency = 10^6, it states that since the gain falls off at a rate of -20dB/dec (see attatchment for graph) starting at w0(Omega-knot) , which is 10^6, the gain will reach 0dB in 2 decades (a factor of 100), therefore unity gain frequency = (10^6)(10^2)= 10^8. Now i don't understand how the gain will reach 0dB in 2 decades, because if you look at the graph of a low pass stc circuit it starts off at 0dB then falls by -20dB/dec. Am i reading the log scale graphs incorrectly? Please clarify. Thank You.

 

[gambit1414]

Sorry, the dc gain (K) is also given which is 100 V/V = 40dB, so it seems like the gain should be 40dB until w=w0 (omega = omega-knot), then the gain starts to attenuate at -20dB/dec. I'm not really good at reading the log graphs so does this mean depending on what your dc gain value is, it will start there and then start decreasing at -20dB/dec after 3dB frequency? Thanks Again.

 

[Mene]

Thank you for your question. I can understand why you may be confused by the solution provided in the manual. Let me try to explain it in more detail.

In a lowpass STC (single time constant) circuit, the gain starts off at 0dB (unity gain) and then decreases as the frequency increases. The rate at which the gain decreases is called the "roll-off rate" and is typically measured in decibels per decade (dB/dec). This means that for every decade increase in frequency, the gain decreases by the specified amount.

In your case, the roll-off rate is -20dB/dec, which means that for every decade increase in frequency, the gain decreases by 20dB. So, at the 3dB frequency (10^6), the gain has decreased by 20dB. This means that at the next decade (10^7), the gain will decrease by another 20dB, making it -40dB. At the next decade (10^8), the gain will decrease by another 20dB, making it -60dB. And so on.

Now, if we want to find the frequency at which the gain becomes 0dB, we need to find the point where the gain reaches unity gain again (0dB). Since the roll-off rate is -20dB/dec, it will take 2 decades (a factor of 100) for the gain to decrease from -40dB to 0dB. This means that the frequency at which the gain becomes 0dB will be 2 decades (a factor of 100) higher than the 3dB frequency, which is why the solution manual states that the unity gain frequency is (10^6)(10^2) = 10^8.

I hope this explanation helps to clarify the concept for you. If you have any further questions, please let me know.