Effect-Site Concentration as a Convolution of Plasma Exposure

Why the Effect Site Remembers Plasma Exposure

An effect-compartment model describes a delay between plasma concentration and the concentration associated with a pharmacologic response. The effect-site concentration does not generally equal the plasma concentration at the same instant: it adjusts toward plasma concentration over time. The convolution representation makes this memory explicit. It expresses the effect-site concentration as a weighted summary of past plasma concentrations, with more recent exposure receiving greater weight.

This representation is useful for understanding what the equilibration rate $K_{e0}$ does. It is not a separate pharmacodynamic mechanism: it is an equivalent way to express the solution to the first-order effect-compartment equation. The response model can then use the resulting effect-site concentration as its concentration input.

Definition: Let $C_p(t)$ be the plasma concentration and $C_e(t)$ the effect-site concentration, both in concentration units. Under the first-order effect-compartment model, $K_{e0}>0$ is the equilibration rate, with units of inverse time, and $$ \frac{dC_e(t)}{dt}=K_{e0}\bigl(C_p(t)-C_e(t)\bigr). $$ The convolution solution from a starting time $t_0$, given an initial effect-site concentration $C_e(t_0)$, is $$ C_e(t)=C_e(t_0)e^{-K_{e0}(t-t_0)} +K_{e0}\int_{t_0}^{t}e^{-K_{e0}(t-s)}C_p(s)\,ds,\qquad t\ge t_0. $$

Deriving the Convolution

Assume $K_{e0}$ is constant over the interval, and that $C_p(t)$ is a known, integrable concentration history. The first-order equation is linear in $C_e(t)$. Multiply both sides by the integrating factor $e^{K_{e0}t}$. By the product rule, the left-hand side becomes the derivative of $e^{K_{e0}t}C_e(t)$:

e^{K_{e0}t}\frac{dC_e(t)}{dt} +K_{e0}e^{K_{e0}t}C_e(t) =\frac{d}{dt}\left(e^{K_{e0}t}C_e(t)\right) =K_{e0}e^{K_{e0}t}C_p(t).

Integrate from $t_0$ to $t$, then divide by $e^{K_{e0}t}$:

e^{K_{e0}t}C_e(t)-e^{K_{e0}t_0}C_e(t_0) =K_{e0}\int_{t_0}^{t}e^{K_{e0}s}C_p(s)\,ds, $$ $$ C_e(t)=C_e(t_0)e^{-K_{e0}(t-t_0)} +K_{e0}\int_{t_0}^{t}e^{-K_{e0}(t-s)}C_p(s)\,ds.

The first term is the surviving contribution of the initial condition. The integral is the accumulated contribution of plasma concentrations between $t_0$ and $t$. For an exposure at time $s$, its lag at observation time $t$ is $u=t-s$, so the kernel is $K_{e0}e^{-K_{e0}u}$. The kernel is largest at lag zero and decreases exponentially as the exposure becomes older.

The kernel has units of inverse time, so multiplying it by concentration and integrating over time produces concentration units. Its total weight over the observed interval is $1-e^{-K_{e0}(t-t_0)}$, not necessarily one. The remaining weight, $e^{-K_{e0}(t-t_0)}$, belongs to the initial concentration. Thus, if the initial concentration and plasma concentrations are nonnegative, the solution is a weighted combination of those concentrations, with weights that sum to one.

What Ke0 Changes About the Weighting

For two exposure times with lags $u_2>u_1\ge 0$, their kernel weights have ratio

\frac{K_{e0}e^{-K_{e0}u_2}}{K_{e0}e^{-K_{e0}u_1}} =e^{-K_{e0}(u_2-u_1)}.

This ratio shows how quickly the relative influence of older exposure falls. A larger $K_{e0}$ makes the kernel more concentrated near the present, so the effect site tracks recent plasma concentrations more closely. A smaller $K_{e0}$ spreads the weighting across a longer history, producing more smoothing and a more persistent influence of earlier exposure. $K_{e0}$ changes the temporal weighting; it does not change the concentration units or directly specify the magnitude of the downstream response.

Worked Example: Constant Plasma Concentration as a Weighted History

Suppose, hypothetically, that $C_e(0)=0$, $K_{e0}=0.30\ \text{h}^{-1}$, and $C_p(s)=8.0\ \text{mg/L}$ from $s=0$ through $t=2.0\ \text{h}$. Find $C_e(2.0\ \text{h})$ from the convolution.

The initial-condition term is zero. Substituting the constant plasma concentration into the integral gives

C_e(2)=0.30\int_0^2 e^{-0.30(2-s)}(8.0)\,ds =8.0\left(1-e^{-0.30(2)}\right).

Since $e^{-0.60}=0.5488$,

C_e(2)=8.0(1-0.5488) =8.0(0.4512) =3.610\ \text{mg/L}.

The result is below the constant plasma concentration of $8.0\ \text{mg/L}$, because the effect site began at zero and had only two hours to respond. The convolution assigns some weight to the initial state and the rest to the plasma exposure during that interval.

Worked Example: Contributions From Two Exposure Periods

Consider a hypothetical profile with $C_e(0)=0$ and $K_{e0}=0.50\ \text{h}^{-1}$. Let $C_p(s)=10\ \text{mg/L}$ from 0 to 1 hour and $C_p(s)=4.0\ \text{mg/L}$ from 1 to 3 hours. Calculate $C_e(3\ \text{h})$, separating the two contributions.

For a constant concentration $C$ over an interval $[a,b]$, its contribution at time $t$ is

K_{e0}C\int_a^b e^{-K_{e0}(t-s)}\,ds =C\left(e^{-K_{e0}(t-b)}-e^{-K_{e0}(t-a)}\right).

For the first interval, the concentration is $10\ \text{mg/L}$, $a=0$, and $b=1$:

10\left(e^{-0.50(3-1)}-e^{-0.50(3-0)}\right) =10(e^{-1}-e^{-1.5}) =10(0.3679-0.2231) =1.448\ \text{mg/L}.

For the second interval, the concentration is $4.0\ \text{mg/L}$, $a=1$, and $b=3$:

4.0\left(e^{-0.50(3-3)}-e^{-0.50(3-1)}\right) =4.0(1-e^{-1}) =4.0(1-0.3679) =2.528\ \text{mg/L}.

Adding the contributions gives $C_e(3)=1.448+2.528=3.976\ \text{mg/L}$. Although the earlier interval had a higher plasma concentration, it contributes less at time 3 because its exposure is older. Each interval’s contribution depends on both its concentration and its position in time.

Worked Example: Same Current Plasma Concentration, Different Histories

Two hypothetical profiles both have $C_p(2)=5.0\ \text{mg/L}$, and both have $K_{e0}=0.40\ \text{h}^{-1}$ and $C_e(0)=0$. In Profile A, plasma concentration is $0\ \text{mg/L}$ from 0 to 1 hour and $5.0\ \text{mg/L}$ from 1 to 2 hours. In Profile B, it is $10\ \text{mg/L}$ from 0 to 1 hour and $5.0\ \text{mg/L}$ from 1 to 2 hours. Compare their effect-site concentrations at 2 hours.

The common contribution from the final hour is

5.0(1-e^{-0.40(1)}) =5.0(1-0.6703) =1.648\ \text{mg/L}.

Profile A has no contribution from the first hour, so $C_{e,A}(2)=1.648\ \text{mg/L}$. Profile B has an additional contribution from its earlier $10\ \text{mg/L}$ concentration:

10\left(e^{-0.40(2-1)}-e^{-0.40(2-0)}\right) =10(e^{-0.4}-e^{-0.8}) =10(0.6703-0.4493) =2.210\ \text{mg/L}.

Therefore $C_{e,B}(2)=1.648+2.210=3.858\ \text{mg/L}$. The current plasma concentrations match, but the effect-site concentrations do not. The convolution retains information about prior exposure that is absent from the single current value $C_p(2)$.

Assumptions, Diagnostics, and Alternatives

The convolution is exact for the stated first-order model when $K_{e0}$ is constant and the plasma history and initial condition are specified. It does not require plasma concentration to be constant; it requires that the history be available in a form that can be integrated. In applications, the plasma profile may be measured or predicted, and the integral may need to be evaluated from that profile. The mathematical formula alone does not establish that the effect site is a physically measured compartment.

A useful basic check is whether the result is consistent with the inputs. With nonnegative initial and plasma concentrations, the weighted-combination interpretation implies $C_e(t)$ lies between the smallest and largest concentrations contributing over the interval. A result outside that range suggests an arithmetic, unit, or implementation error. Other warning signs include an effect-site curve that responds in the wrong temporal direction, implausible dependence on the initial condition, or systematic disagreement between predicted responses and observed data. Such disagreement may reflect incorrect plasma input, a poor choice of $K_{e0}$, or inadequacy of the first-order effect-compartment structure; the convolution does not diagnose which explanation is responsible.

An instantaneous-effect assumption uses $C_p(t)$ directly and discards exposure history. It can be a reasonable approximation when equilibration is very rapid relative to changes in plasma concentration, but it cannot represent substantial temporal lag. A fixed-delay model instead shifts the input in time; unlike the exponential convolution, it does not blend contributions over a range of past times. More elaborate dynamic models can represent multiple stages or other patterns of delay, but they add parameters and assumptions. The first-order convolution is attractive because it has a clear interpretation and a compact description, not because it is guaranteed to fit every delayed response.

A common mistake is to call $C_e(t)$ a simple average of all past plasma concentrations. It is not: older values are downweighted exponentially, the initial condition can contribute, and the total weight assigned to observed plasma history depends on elapsed time. Another mistake is to interpret a larger $K_{e0}$ as a larger pharmacologic effect. In this model, $K_{e0}$ controls equilibration speed; the concentration-effect relationship determines how concentration maps to response.

Key takeaway: The effect-site concentration is the initial effect-site state plus an exponentially weighted integral of plasma exposure. $K_{e0}$ determines how rapidly the weights decline with exposure age: larger values emphasize recent plasma concentrations, while smaller values retain more influence from earlier exposure.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

In the convolution formula, what does the initial-condition term represent, and why can it not always be omitted?
For two exposure times separated by 2 hours, what is the ratio of the older time’s kernel weight to the more recent time’s weight when $K_{e0}=0.25\ \text{h}^{-1}$?
Suppose $C_e(0)=0$, $K_{e0}=0.20\ \text{h}^{-1}$, and plasma concentration is $6.0\ \text{mg/L}$ for 1 hour. Calculate the effect-site concentration at 1 hour.
Two profiles have the same plasma concentration at the observation time but different earlier concentrations. Can their effect-site concentrations differ? Explain using the convolution.
Name one reason that a first-order convolution may not adequately describe an observed delayed response.

Solutions. (1) It is the initial effect-site concentration multiplied by the fraction that remains after elapsed time $t-t_0$. It can be omitted only when that contribution is zero, such as when $C_e(t_0)=0$. (2) The ratio is $e^{-K_{e0}(2)}=e^{-0.50}=0.6065$; the older exposure has about 60.7% of the kernel weight of the more recent one. (3) $C_e(1)=6.0(1-e^{-0.20})=6.0(1-0.8187)=1.088\ \text{mg/L}$. (4) Yes. The integral includes the full plasma history, so differing earlier concentrations can contribute differently even when the current values match. (5) For example, the true delay may involve multiple dynamic stages rather than one first-order equilibration process, or the plasma input may be misspecified.

Effect-Site Concentration as a Convolution of Plasma Exposure

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