If you set V(t ¼ 0) ¼ 0 and W(t ¼ 0) ¼ 0, both the membrane potential and
recovery mechanism are at equilibrium. If you further set E ¼ 0, the cell does not
respond because all potentials are zero.
When the neuron is exposed to either a dendritic voltage or an external potential,
the cell membrane potential builds and “fires” an amplification of the combined
received voltages. The dendritic signals are simulated by initial values of the
membrane potential. The membrane potential response shows a critical threshold
voltage potential is required. A quick examination of the differential equations
shows that this amplification threshold is 0.2 (V1) and amplification ceases when
the voltage exceeds 1.0 (V2)—our choice of parameters for the model.
When the dendritic potential is zero (V(t ¼ 0) ¼ 0, W(t ¼ 0) ¼ 0), the effect of
applied voltages (E) can be seen. For currents over 0.23, the membrane potential
cycles up and down. This result shows an unstable condition that also can be
observed in physical experiments. The results for E ¼ 0.23 are shown in Fig. 11.3.
This is the lowest applied voltage that will produce cycling. Can you find the value
for E that leads to the largest amplification of the input signal?
Figure 11.4 the case of amplification of the dendritic signal: E ¼ 0, V(t ¼ 0) ¼
0.4 and all other variables the same as above. The cell responds when the initial
membrane potential is set to some positive potential. The membrane potential
asymptotically approaches zero from its initial value, after rising, then overshooting
zero and dropping below the equilibrium potential (zero on our scale). Try
EPSILON ¼ 0.002 and note the different level of response. Demonstrate that the
amplification range of the cell is actually between 0.2 V(t ¼ 0) 1.0. Why do you
suppose the nerve cell amplifies the combination of the input signals?
The first model run above showed cycling of V and W for E ! 0.23. Keep
increasing EPSILON for subsequent runs and observe the results. You will find
Fig. 11.2
11.1 Fitzhugh–Nagumo Neuron Model
93
recovery mechanism are at equilibrium. If you further set E ¼ 0, the cell does not
respond because all potentials are zero.
When the neuron is exposed to either a dendritic voltage or an external potential,
the cell membrane potential builds and “fires” an amplification of the combined
received voltages. The dendritic signals are simulated by initial values of the
membrane potential. The membrane potential response shows a critical threshold
voltage potential is required. A quick examination of the differential equations
shows that this amplification threshold is 0.2 (V1) and amplification ceases when
the voltage exceeds 1.0 (V2)—our choice of parameters for the model.
When the dendritic potential is zero (V(t ¼ 0) ¼ 0, W(t ¼ 0) ¼ 0), the effect of
applied voltages (E) can be seen. For currents over 0.23, the membrane potential
cycles up and down. This result shows an unstable condition that also can be
observed in physical experiments. The results for E ¼ 0.23 are shown in Fig. 11.3.
This is the lowest applied voltage that will produce cycling. Can you find the value
for E that leads to the largest amplification of the input signal?
Figure 11.4 the case of amplification of the dendritic signal: E ¼ 0, V(t ¼ 0) ¼
0.4 and all other variables the same as above. The cell responds when the initial
membrane potential is set to some positive potential. The membrane potential
asymptotically approaches zero from its initial value, after rising, then overshooting
zero and dropping below the equilibrium potential (zero on our scale). Try
EPSILON ¼ 0.002 and note the different level of response. Demonstrate that the
amplification range of the cell is actually between 0.2 V(t ¼ 0) 1.0. Why do you
suppose the nerve cell amplifies the combination of the input signals?
The first model run above showed cycling of V and W for E ! 0.23. Keep
increasing EPSILON for subsequent runs and observe the results. You will find
Fig. 11.2
11.1 Fitzhugh–Nagumo Neuron Model
93