The variable $I_d$ is an approximation to the value of $I$ delayed by $\delta$, where $\delta$ is measured in ms. In the following, we will vary the strength $a$ of the inhibitory feedback and examine how the delay required for an oscillation changes.
For all numerical answers, include at least 4 significant digits.
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A preliminary check, let's review the system without a delay to make sure we cannot get an oscillation without a delay. If we eliminate the delay, setting $I_d$ to $I$ so that the system becomes
\begin{align*}
\diff{E}{t} &= \frac{1}{\tau_e}(-E - a I + P)\\
\diff{I}{t} &= \frac{1}{\tau_i}(-I + bE+Q),\\
\end{align*}
can the system oscillate for any value of $a$?
, because no what the value of $a$ is, the trace of the matrix underlying the system is
and the determinant is
. The equilibrium is always
and can never be a
.
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Restoring the delay, so that we are back to the system of three equations involving $I_d$, let's first examine the case when $a=4$.
For $a=4$, find two values of the delay $\delta$ for which the system oscillates.
$\delta_1 = $
ms, $\delta_2 = $
ms
You should have found one value less than 100 ms and the other value greater than 100 ms. We will focus on the shorter delay, as the longer delay is much too long to be realistic for delays in a local neuronal circuit. The shorter delay is $\delta=$
ms.
When the delay is tuned to this value, what is the frequency of the oscillations?
Hz.
Hint
The simplest way to calculate the required delay is to use the Routh-Horwitz Matlab script included with the textbook. However that script defaults to Matlab's short format with a format short
command at the beginning. You can change that command to format long
to see more digits.
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For $a=8$, find the value of the delay $\delta$ that is less than 100 ms for which the system oscillates.
$\delta = $
ms
When the delay is tuned to this value, what is the frequency of the oscillations?
Hz.
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For $a=15$, find the value of the delay $\delta$ that is less than 100 ms for which the system oscillates.
$\delta = $
ms
When the delay is tuned to this value, what is the frequency of the oscillations?
Hz.
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As the inhibitory feedback becomes stronger, we observe the following trends. The length of the delay required for oscillations becomes
and the oscillation frequency becomes
.