| Chronology | Current Month | Current Thread | Current Date |
| [Year List] [Month List (current year)] | [Date Index] [Thread Index] | [Thread Prev] [Thread Next] | [Date Prev] [Date Next] |
At 04:28 PM 4/9/02, Paul Giusti wrote:cut-off
... a small question i cannot do. it is
on analogue electronics...
an amplifier has a second order gain/frequency response with a lower
asfrequency of 350Hz, an upper cut-off frequency of 4kHz and a mid-band
voltage of 50 dB.
sketch the graph of gain against frequency.
estimate the voltage gain in dB at the following frequencies.
i) 20Hz
ii) 350Hz
iii) 2kHz
iv) 10kHz
thanks if any of you can help.
paul
An RC circuit has the property that the output voltage halves when the
frequency doubles.
As you will know, the R element holds a constant value, but the C element
has a reactance proportional to 1/fC.
A doubling is labeled as an octave, and a voltage halving is described
-6 dB on this logarithmic scale. So a single pole filter is said to fallat
off at -6dB/octave.
A two pole filter falls at -12 dB/octave. This is also called a second
order response.
If I sketch a pass band gain of 50 dB, with a roll off at -12 dB to 0 dB
4kHz, and a slope of 12 dB from 0 dB at 350 Hz, I find I am sketching the
impossible.
It could just be that the corner frequencies specified are meant to be the
pole frequencies at which the slopes begin.
If I suppose the lower pole of 350 Hz represents 50 dB voltage gain then
I might expect 0 dB at 22 Hz, and the high cutoff of 0dB at 64 kHz.
If this were the case, the gain at 20 Hz would be 0dB, at 350 Hz = 50
dB, 2 kHz = 50 dB, 10 kHz = 32dB
I am however doubtful that I have correctly interpreted the intent of this
question.
Brian Whatcott
Altus OK Eureka!