Concept of Transient Response of Electrodes

Transient Specification:

In simple analogue amplifiers and interfaces it is practically not possible to obtain the linear phase characteristic that is needed to prevent distortion of a signal. In an attempt to cater for this, early standards for electrocardiographs gave a sign of the phase distortion that was acceptable. They did this by stipulating the phase shift introduced to the signal by recording system must be no more than that introduced by a single-pole, high-pass filter containing a cut-off frequency of 0.05Hz. This left some ambiguity surrounding the consequence this would have on an ECG signal that is usually viewed in the time domain for diagnostic purposes. In an effort to quantify the distortion in time domain more recent standards have pointed this in terms of transient response to a narrow pulse as shown in figure below. The distortion is specified in terms of maximum undershoot from the baseline at the end of pulse that can be tolerated and the maximum recovery slope from this undershoot is permissible. The specification is provided for an input rectangular pulse of amplitude 3mV and duration 100ms. The maximum undershoot from the baseline at the end of pulse is limited to 100μV, whereas the maximum recovery slope is limited to 300μVs-1. Such values apply to the total distortion in the ECG introduced at any point in the recording system and thus should be taken as being termed to the input and hence they are independent of the variable amplifier gain.

Figure: The Transient Response Specification for ECG Recorders

Transient Analysis:

The similar model can be employed as before for the electrode amplifier interface as shown in figure below, with the sinusoidal source substituted with a pulse generator intended to offer the pulse shown in figure above.

Figure: A Model of the Front End of the Recoding System

Once again, the input impedance of the amplifier is taken as purely resistive and the polarisation potential is ignored. Formerly the steady-state response of this set-up was explained in the frequency domain as:

Vin/Vs = α [{1 + j(ω/ωZ)}]/ [{1 + j(ω/ωP)}]

In order to assess the transient response the term ‘jω’ is substituted by the Laplace operator‘s’ and hence:

Vin(S)/Vs(S) = α [{1 + (S/Z)}]/ [{1 + (S/P)}]


Z = 1/(CP RP)
P = 1/(α CP RP) = Z/α
α = Rin/[Rin + 2(RS + RP)]

And hence,

Vin(S)/VS(S) = α [(1/z)(S+z)]/[(1/p)(S+P)] = α {P(S+z)}/{z(S+P)} = (S+z)/(S+P)

When the input pulse has amplitude Vm and a duration T as shown in figure below:

Figure: Input Test Pulse

This can be conveniently defined as a sum of two step functions and hence:

VS(t) Vm u(t) – Vm u (t - T)

This consists of the Laplace transform:

VS(S) = (Vm/S) - (Vm/S) e-ST = (Vm/S) (1 - e-ST)

The input voltage to amplifier is then given in the Laplace terms as the product of transform of the input pulse and transfer-function of the electrode-amplifier interface and hence:

Vin(S) = VS(S) [Vin(S)/VS(s)]

This gives:

Vin(S) = (Vm/s) (1 – e-ST) [(S + Z)/(S+P)]

The appropriate polynomial fraction in the Laplace operator‘s can be simplified by the partial fraction expansion:

(S + Z)/S(S + P) = (A/S) + B/(S + P)

(S + Z)/S(S + P) = [A(S + P) + BS]/ S(S + P)

(S + Z)/S(S + P) = [(A + B)S + AP]/S(S + P)

On comparing the coefficients gives:

A + B = 1; AP =Z

A = Z/P = α Z/Z = α; B = 1 – A = 1 – α

And hence,

(S + Z)/S(S + P) = (α/S) + (1 - α)/(S + P)

The voltage at input to the amplifier is given in Laplace form as follows:

Vin (S) = Vm (1 – e-ST) [(α/S) + (1 - α)/(S + P)]

This expands to:

Vin (S) = [Vm (α/S)] – [Vm (α/S) (e-ST)] + [Vm {(1 - α)/ (S + P)}] – [Vm {(1 - α)/ (S + P)} (e-ST)]

By taking the inverse Laplace transform provides:

Vin(t) = Vm α u (t) - Vm α u (t - T) + Vm (1 - α) (e-pt) u(t) – Vm (1- α) (e-p(t-T)) u(t - T)

That simplifies to:

Vin(t) = Vm [{α + (1- α) (e-pt) }u(t) – { α + (1- α) (e-p(t-T))}u(t - T)]

Whenever this expression is examined it can be observe to agree with the response shown in figure above to the input rectangular pulse whereby the pulse whenever present experiences some ‘droop’ in its transfer to amplifier input, then some undershoot of baseline at the end of pulse, ultimately followed by an exponential recovery to baseline.

If t= 0 we have,

Vin (t = 0) = Vm [{α + (1- α) (eo)}1 - [{α + (1- α) (epT)}0] = Vm

There is no preliminary attenuation of the pulse as we took RS << Rin formerly, that essentially ignores the small amount of attenuation caused by the RS.

If 0 ≤ t < T we encompass:

Vin(t) = Vm [{α + (1- α) (e-pT)}]

This begins at a value of Vm when t = 0 and decays exponentially towards the value of αVm. The exponential decay is governed by pole time constant α CP RP. In practice, the voltage never reaches its final value as time constant related with the electrode components is much bigger than the width of the pulse T of 100ms.

When t = T then, at the end of pulse:

Vin(t = T) = Vm[{α + (1- α) (e-pT)} - {α + (1- α) (e-0)1}]

Vin(t = T) = Vm [{α + (1- α) (e-pT)}- 1]

Vin(t = T) = Vm (1 - α) (e-pT) - Vm + αVm

Vin(t = T) = Vm (1 - α) (e-pT) - Vm (1 - α)

Vin(t = T) = - [(1 - α)(1 - e-pT)] Vm

This explains the value of pulse with the level of droop prevailing at the end of pulse at t = T with a subtraction of Vm volts as the pulse terminates. This provides the value of undershoot from baseline at this time.

Finally, if t > T then the voltage at the amplifier input is explained as:

Vin(t) = Vm [(1 - α) (e-pt) - (1- α) (e-p(t-T))]

The manipulation gives:

Vin(t) = (1 - α)Vm [e-pt - e-p(t-T)]

Vin(t) = [(1 - α)(1 - epT)] Vme-pt

Vin(t) = [(1 - α)(1 - epT)] Vme-pt e-pt e-pt ept

Vin(t) = [(1 - α)(e-pT - 1)] Vme-pt ept

Vin(t) = - [(1 - α)(e-pT - 1)] Vme-p(t-T)

This explains the exponential recovery of undershoot from the baseline that is again governed by the pole time constant.

Maximum Undershoot Limitation:

The value of undershoot from baseline at the end of pulse is given as:

Vin (t = T) = - [(1 - α) (e-pt)] Vm

This can be stated as a fraction of the pulse amplitude merely as:

[Vin (t = T)]/Vm = - [(1 - α) (e-pt)]

The specification shown in first figure gives the undershoot limit as 100μV for an input pulse amplitude 3mV. This corresponds to the fractional value of 0.033 or 3.3%. In order to fulfill this need:

(1 - α) (e-pt) < 0.033

Though, p = z/α and hence:

(1 - α) (e-(ZT/α)) < 0.033

Power series expansion for exponential is as follows:

ex = 1 + x + (x2/2!) + (x3/3!) + (x4/4!) + .............

By using just the first order terms, the above inequality approximates to:

(1 - α) (ZT/α) < 0.033

[(1 - α)/α] ZT < 0.033

[(1 - α)/α] [T/CP RP] < 0.033

General values for the electrode components for modern disposable, adhesive electrodes have been provided formerly as RS = 50Ω, RP = 200kΩ and CP = 0.5µF.

With T = 100ms = 10-5 s we encompass:


Most of the ECG amplifiers use an input impedance of 10MΩ that is recommended in AHA standards. For the values of equivalent electrical components provided, this is not enough to meet the utmost undershoot limitation. The value closer to 20MΩ would be required to make sure that the need is met. Electrodes with higher values of components will require even higher amplifier input impedance.

Maximum Recovery of Slope Limitation:

The profile of exponential recovery of the voltage after undershoots at the end of pulse at amplifier input is explained by:

Vin(t) = - [(1 - α) (1 - e-pT)] Vm e-p(t -T)

Slope of this profile is as follows:

dVin(t)/dt = [(1 - α) (1 - e-pT)] Vm Pe-p(t –T)

If t = T this has a value:

dVin(t)/dt |t = T = [(1 - α) (1 - e-pT)] PVm

When the exponential term above is approximated by first order terms of an expansion as prior to the expression becomes:

dVin(t)/dt |t = T = (1 - α) P2 TVm

With p = z/α = 1/α CP RP this becomes:

dVin(t)/dt |t = T = [(1 - α)/{α2 (CPRP)2 }TVm]

The pulse defined in first figure has the properties Vm = 3mV and T = 100ms and having values of the electrode components as before, RP = 200kΩ and CP = 0.5µF:


Limit of the recovery slope as specified in first figure is given as 300μVs-1.This needs:

[(1 - α)/α2] x 3 x 10-2 < 3 x 10-4

And hence,

[(1 - α)/α2] < 10-2

Inverting gives:

α2/(1 - α) > 102

This needs a value of α that is very close to unity. In this case α → 1 and hence α2 ≈ α. Then the need can be approximated as:

α/(1 - α) > 102

α > (1 - α) x 102

α (1 +2) > 102

α > 100/101 > 0.99

This signifies:

α = Rin/{Rin + 2(RS + RP)} > 0.99

Rin > 0.99 Rin + 1.98 (RS + RP)

0.01 Rin > 1.98 (RS + RP)

Rin >198 (RS + RP)

Using the values of RS = 50Ω, RP = 200kΩ as before this signifies:

Rin > 39.6 MΩ

This places noteworthy demands on the design of amplifier input phase.

Latest technology based Electrical Engineering Online Tutoring Assistance

Tutors, at the, take pledge to provide full satisfaction and assurance in Electrical Engineering help via online tutoring. Students are getting 100% satisfaction by online tutors across the globe. Here you can get homework help for Electrical Engineering, project ideas and tutorials. We provide email based Electrical Engineering help. You can join us to ask queries 24x7 with live, experienced and qualified online tutors specialized in Electrical Engineering. Through Online Tutoring, you would be able to complete your homework or assignments at your home. Tutors at the TutorsGlobe are committed to provide the best quality online tutoring assistance for Electrical Engineering Homework help and assignment help services. They use their experience, as they have solved thousands of the Electrical Engineering assignments, which may help you to solve your complex issues of Electrical Engineering. TutorsGlobe assure for the best quality compliance to your homework. Compromise with quality is not in our dictionary. If we feel that we are not able to provide the homework help as per the deadline or given instruction by the student, we refund the money of the student without any delay.