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Wave shaping circuitsa`
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High Pass Circuit
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High Pass Circuit
Signal is attenuated or damped at low frequencies with the output
increasing at +20dB/Decade (6dB/Octave) until the frequency reaches
the cut-off point ( c) where again R = Xc.
At cut-off frequency, where the output voltage amplitude
is 1/2 = 70.7% of the input signal value or -3dB (20 log (Vout/Vin)) ofthe input value.
Phase angle ( ) of the output signal LEADSthat of the input and isequal to +45oat frequency c.
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Step Response
Voltage on capacitor cannot change instantaneously. So Vout = Vin initially.
Response of RC High Pass Circuit to Standard waveforms
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RC
Fall Time
Vout
time
1.0
0.9
0.1 10%
90%
100%
1/e~37%
e
t
RC
Fall Time & Time Constant ( )
Response of RC High Pass Circuit to Standard waveforms
Step Response
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Response of RC High Pass Circuit to Standard waveforms
Pulse Response
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Response of RC High Pass Circuit to Standard waveforms
Square Wave Response Response
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Response of RC High Pass Circuit to Standard waveforms
Square Wave Response Response
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Response of RC High Pass Circuit to Standard waveforms
Ramp Response
Let a, be the slop of input ramp signal therefore it can be representedas : Vi(t) = at Now Apply KVL to RC circuit.
Vi = Vc + VR
Vi = (q/c) + VR ---- 1
at = (q/c) + Vo ---- 2Differentiate eq. 2 on both sides wrt t we get
--3
Since dq/dt = i Vo = i . R i = Vo/R = dq/dt = Vo/R
Therefore , -- 4
Hence solution to this differential equation is given by,
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Response of RC High Pass Circuit to Standard waveforms
Ramp Response
]1[ RCt
eaRCVo
Nature of output depends on Value of RC time constant.
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Response of RC High Pass Circuit to Standard waveforms
High-pass RC circuit as Differentiator:
A circuit in which the output voltage is directly proportional to the
derivative of the input voltage is called a differentiating circuit.
Mathematically, the output voltage is given by:
Output (Vo) d/dt input (Vi)
If a d.c. or constant input is applied to such a circuit, the output will bezero. It is because the derivative of the constant is zero.
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Response of RC High Pass Circuit to Standard waveforms
High-pass RC circuit as Differentiator:
Let Vi , be the input alternating voltage and let i be The resultingalternating current. The charge q on the capacitor at any instant is:
q = C.Vc and i = dq/dt
i = d/dt (CVc)
i = C d/dt (Vc)
Since the capacitive reactance is very larger than R, the input voltage
can be consider equal to the capacitor voltage without any error,, i.e
Vc= Vi ,
i = C . d/dt (Vi)
output voltage is given by:Vo= iR Or Vo=(C . d/dt (Vi)).R
Vo=RC d/dt ( Vi)
where RC is a constant, and Hence Output d/dt (input)
RC High pass circuit can work as differentiator for smaller values ofRC time constant.
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Response of RC High Pass Circuit to Standard waveforms
Applications:Some important applications of a differentiating circuit are given as
under:
To generate a square wave from a triangular wave input.
To generate a step from a ramp input.
To generate a series of narrow pulses called spikes from the
rectangular or square waveform. The pips are used as trigger pulses or
synchronization pulses in circuits used in television and cathode ray
oscilloscopes.
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Low Pass Circuit
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Low Pass Circuit
This Cut-off, Corner or Breakpoint frequency is defined as
being the frequency point where the capacitive reactance and resistanceare equal, R = Xc . When this occurs the output signal is attenuated
to 70.7% of the input signal value or -3dB (20 log (Vout/Vin)) of the
input.
The Phase Angle ( ) of the output signal LAGSbehind that of theinput and at the -3dB cut-off frequency ( c) and is - 45oout of phase.
This is due to the time taken to charge the plates of the capacitor as the
input voltage changes, resulting in the output voltage (the voltage acrossthe capacitor) laggingbehind that of the input signal.
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