2007-02-11 06:28:50 · 6 answers · asked by naurinbutt 1 in Science & Mathematics ➔ Engineering
the transient time for the RC circuit used is high enough that the time needed for the high freq signal to change from peak to peak is much higher than the time of the transient time for the RC filter, which in turn removes the high freq.
2007-02-11 08:35:54 · answer #1 · answered by Anonymous · 0⤊ 0⤋
This circuit is quite a bit more complicated than a parallel resistor capacitor circuit. For example R1 R2 and C2 form a passive high pass filter. T1 and T2 amplify this signal from the microphone after it has been filtered (most likely to remove frequencies above and below the vocal range). Finally the right hand part of the circuit combines this lower frequency "voice" signal with the much higher frequency RF signal and is amplified by T3 and output to the antenna. Within the circuit R11 and C6 would be an example of a parallel RC network. Hope that helps!
2016-05-23 21:56:25 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Look at the bode plot of a low pass filter. After the 1/RC frequency the magnitude of the signal starts decreasing by 20 dB/dec if it is a single pole at that frequency. Double poles are 40 dB/dec, triple are 60 and so on.
Check the link below.
2007-02-11 14:58:17 · answer #3 · answered by Spartian 1 · 0⤊ 0⤋
It's a voltage divider.
There's the R in series with the source, then the C to "ground" or circuit common. The output is taken form the junction of the two.
As the applied frequency rises Xc reduces (but R stays the same), so the voltage divider action reduces the amplitude of the output.
2007-02-11 13:53:25 · answer #4 · answered by dmb06851 7 · 0⤊ 0⤋
Because it's a LOW PASS filter.
2007-02-11 06:42:58 · answer #5 · answered by Anonymous · 0⤊ 1⤋
cause its a low pass
2007-02-11 06:33:47 · answer #6 · answered by wade 3 · 0⤊ 1⤋