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# Configure Jupyter so figures appear in the notebook
%matplotlib inline
# Configure Jupyter to display the assigned value after an assignment
%config InteractiveShell.ast_node_interactivity='last_expr_or_assign'
# import functions from the modsim.py module
from modsim import *
The following circuit diagram (from Wikipedia) shows a low-pass filter built with one resistor and one capacitor.
A "filter" is a circuit takes a signal, $V_{in}$, as input and produces a signal, $V_{out}$, as output. In this context, a "signal" is a voltage that changes over time.
A filter is "low-pass" if it allows low-frequency signals to pass from $V_{in}$ to $V_{out}$ unchanged, but it reduces the amplitude of high-frequency signals.
By applying the laws of circuit analysis, we can derive a differential equation that describes the behavior of this system. By solving the differential equation, we can predict the effect of this circuit on any input signal.
Suppose we are given $V_{in}$ and $V_{out}$ at a particular instant in time. By Ohm's law, which is a simple model of the behavior of resistors, the instantaneous current through the resistor is:
$ I_R = (V_{in} - V_{out}) / R $
where $R$ is resistance in ohms.
Assuming that no current flows through the output of the circuit, Kirchhoff's current law implies that the current through the capacitor is:
$ I_C = I_R $
According to a simple model of the behavior of capacitors, current through the capacitor causes a change in the voltage across the capacitor:
$ I_C = C \frac{d V_{out}}{dt} $
where $C$ is capacitance in farads (F).
Combining these equations yields a differential equation for $V_{out}$:
$ \frac{d }{dt} V_{out} = \frac{V_{in} - V_{out}}{R C} $
Follow the instructions blow to simulate the low-pass filter for input signals like this:
$ V_{in}(t) = A \cos (2 \pi f t) $
where $A$ is the amplitude of the input signal, say 5 V, and $f$ is the frequency of the signal in Hz.
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ohm = UNITS.ohm
farad = UNITS.farad
volt = UNITS.volt
Hz = UNITS.Hz
second = UNITS.second
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params = Params(
R1 = 1e6 * ohm,
C1 = 1e-9 * farad,
A = 5 * volt,
f = 1000 * Hz)
Now we can pass the Params object make_system which computes some additional parameters and defines init.
omega is the frequency of the input signal in radians/second.
tau is the time constant for this circuit, which is the time it takes to get from an initial startup phase to
cutoff is the cutoff frequency for this circuit (in Hz), which marks the transition from low frequency signals, which pass through the filter unchanged, to high frequency signals, which are attenuated.
t_end is chosen so we run the simulation for 4 cycles of the input signal.
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def make_system(params):
"""Makes a System object for the given conditions.
params: Params object
returns: System object
"""
f, R1, C1 = params.f, params.R1, params.C1
init = State(V_out = 0)
omega = 2 * np.pi * f
tau = R1 * C1
cutoff = 1 / R1 / C1 / 2 / np.pi
t_end = 4 / f
dt = t_end / 4000
return System(params, init=init, t_end=t_end, dt=dt,
omega=omega, tau=tau, cutoff=cutoff.to(Hz))
Let's make a System
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system = make_system(params)
Exercise: Write a slope function that takes as an input a State object that contains V_out, and returns the derivative of V_out.
Note: The ODE solver requires the return value from slope_func to be a sequence, even if there is only one element. The simplest way to do that is to return a list with a single element:
return [dV_out_dt]
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# Solution goes here
Test the slope function with the initial conditions.
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slope_func(system.init, 0*UNITS.s, system)
And then run the simulation. I suggest using t_eval=ts to make sure we have enough data points to plot and analyze the results.
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results, details = run_ode_solver(system, slope_func)
details
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results.head()
Here's a function you can use to plot V_out as a function of time.
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def plot_results(results):
xs = results.V_out.index
ys = results.V_out.values
t_end = get_last_label(results)
if t_end < 10:
xs *= 1000
xlabel = 'Time (ms)'
else:
xlabel = 'Time (s)'
plot(xs, ys)
decorate(xlabel=xlabel,
ylabel='$V_{out}$ (volt)',
legend=False)
plot_results(results)
If things have gone according to plan, the amplitude of the output signal should be about 0.8 V.
Also, you might notice that it takes a few cycles for the signal to get to the full amplitude.
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fs = [1, 10, 100, 1000, 10000, 100000] * Hz
for i, f in enumerate(fs):
system = make_system(Params(params, f=f))
results, details = run_ode_solver(system, slope_func)
subplot(3, 2, i+1)
plot_results(results)
At low frequencies, notice that there is an initial "transient" before the output gets to a steady-state sinusoidal output. The duration of this transient is a small multiple of the time constant, tau, which is 1 ms.
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system = make_system(Params(params, f=1000*Hz))
results, details = run_ode_solver(system, slope_func)
V_out = results.V_out
plot_results(results)
The following function computes V_in as a TimeSeries:
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def compute_vin(results, system):
"""Computes V_in as a TimeSeries.
results: TimeFrame with simulation results
system: System object with A and omega
returns: TimeSeries
"""
A, omega = system.A, system.omega
ts = results.index.values * UNITS.second
V_in = A * np.cos(omega * ts)
return TimeSeries(V_in, results.index, name='V_in')
Here's what the input and output look like. Notice that the output is not just smaller; it is also "out of phase"; that is, the peaks of the output are shifted to the right, relative to the peaks of the input.
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V_in = compute_vin(results, system)
plot(V_out)
plot(V_in)
decorate(xlabel='Time (s)',
ylabel='V (volt)')
The following function estimates the amplitude of a signal by computing half the distance between the min and max.
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def estimate_A(series):
"""Estimate amplitude.
series: TimeSeries
returns: amplitude in volts
"""
return (series.max() - series.min()) / 2
The amplitude of V_in should be near 5 (but not exact because we evaluated it at a finite number of points).
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A_in = estimate_A(V_in)
The amplitude of V_out should be lower.
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A_out = estimate_A(V_out)
And here's the ratio between them.
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ratio = A_out / A_in
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ratio.to_base_units()
Exercise: Encapsulate the code we have so far in a function that takes two TimeSeries objects and returns the ratio between their amplitudes.
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# Solution goes here
And test your function.
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estimate_ratio(V_out, V_in)
The delay between the peak of the input and the peak of the output is call a "phase shift" or "phase offset", usually measured in fractions of a cycle, degrees, or radians.
To estimate the phase offset between two signals, we can use cross-correlation. Here's what the cross-correlation looks like between V_out and V_in:
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corr = correlate(V_out, V_in, mode='same')
corr = TimeSeries(corr, V_in.index)
plot(corr, color='C4')
decorate(xlabel='Lag (s)',
ylabel='Correlation')
The location of the peak in the cross correlation is the estimated shift between the two signals, in seconds.
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peak_time = corr.idxmax() * UNITS.second
We can express the phase offset as a multiple of the period of the input signal:
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period = 1 / system.f
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(peak_time / period).to_reduced_units()
We don't care about whole period offsets, only the fractional part, which we can get using modf:
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frac, whole = np.modf(peak_time / period)
frac = frac.to_reduced_units()
Finally, we can convert from a fraction of a cycle to degrees:
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frac * 360 * UNITS.degree
Exercise: Encapsulate this code in a function that takes two TimeSeries objects and a System object, and returns the phase offset in degrees.
Note: by convention, if the output is shifted to the right, the phase offset is negative.
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# Solution goes here
Test your function.
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estimate_offset(V_out, V_in, system)
Exercise: Write a function that takes as parameters an array of input frequencies and a Params object.
For each input frequency it should run a simulation and use the results to estimate the output ratio (dimensionless) and phase offset (in degrees).
It should return two SweepSeries objects, one for the ratios and one for the offsets.
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# Solution goes here
Run your function with these frequencies.
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fs = 10 ** linspace(0, 4, 9) * Hz
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ratios, offsets = sweep_frequency(fs, params)
We can plot output ratios like this:
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plot(ratios, color='C2', label='output ratio')
decorate(xlabel='Frequency (Hz)',
ylabel='$V_{out} / V_{in}$')
But it is useful and conventional to plot ratios on a log-log scale. The vertical gray line shows the cutoff frequency.
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def plot_ratios(ratios, system):
"""Plot output ratios.
"""
# axvline can't handle a Quantity with units
cutoff = magnitude(system.cutoff)
plt.axvline(cutoff, color='gray', alpha=0.4)
plot(ratios, color='C2', label='output ratio')
decorate(xlabel='Frequency (Hz)',
ylabel='$V_{out} / V_{in}$',
xscale='log', yscale='log')
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plot_ratios(ratios, system)
This plot shows the cutoff behavior more clearly. Below the cutoff, the output ratio is close to 1. Above the cutoff, it drops off linearly, on a log scale, which indicates that output ratios for high frequencies are practically 0.
Here's the plot for phase offset, on a log-x scale:
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def plot_offsets(offsets, system):
"""Plot phase offsets.
"""
# axvline can't handle a Quantity with units
cutoff = magnitude(system.cutoff)
plt.axvline(cutoff, color='gray', alpha=0.4)
plot(offsets, color='C9')
decorate(xlabel='Frequency (Hz)',
ylabel='Phase offset (degree)',
xscale='log')
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plot_offsets(offsets, system)
For low frequencies, the phase offset is near 0. For high frequencies, it approaches 90 degrees.
By analysis we can show that the output ratio for this signal is
$A = \frac{1}{\sqrt{1 + (R C \omega)^2}}$
where $\omega = 2 \pi f$, and the phase offset is
$ \phi = \arctan (- R C \omega)$
Exercise: Write functions that take an array of input frequencies and returns $A(f)$ and $\phi(f)$ as SweepSeries objects. Plot these objects and compare them with the results from the previous section.
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# Solution goes here
Test your function:
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A = output_ratios(fs, system)
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# Solution goes here
Test your function:
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phi = phase_offsets(fs, system)
Plot the theoretical results along with the simulation results and see if they agree.
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plot(A, ':', color='gray')
plot_ratios(ratios, system)
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plot(phi, ':', color='gray')
plot_offsets(offsets, system)
For the phase offsets, there are small differences between the theoretical results and our estimates, but that is probably because it is not easy to estimate phase offsets precisely from numerical results.
Exercise: Consider modifying this notebook to model a first order high-pass filter, a two-stage second-order low-pass filter, or a passive band-pass filter.
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