From Arterial Pressure to Autonomic Response
The arterial baroreflex is a rapid negative-feedback system. Stretch-sensitive receptors in the carotid sinus and aortic arch change their firing as arterial pressure changes. Neural pathways use that signal to adjust autonomic output: a pressure rise generally increases baroreceptor firing, reduces sympathetic drive, and increases parasympathetic (vagal) influence on the heart. A pressure fall generally shifts the response in the opposite direction. The resulting changes in heart rate, cardiac contractility, and vascular tone can oppose the initial pressure disturbance.
For quantitative systems pharmacology, the aim here is to make that chain explicit: pressure deviation produces a sensor signal; the signal drives autonomic states; those states alter cardiovascular outputs. The equations below describe a deliberately reduced model, not a complete representation of neural physiology. They are useful for reasoning about response direction, timing, and feedback near a specified operating point.
The scale of 25 mmHg is a model choice that converts a pressure change into a dimensionless signal; it is not a universal physiological constant. A local, linear mapping from the sensor signal to autonomic targets is
Thus, a pressure decrease gives \(r<0\), a positive sympathetic target, and a negative parasympathetic target. Those signs encode increased sympathetic and reduced vagal drive. A pressure increase gives the reverse. The gains are illustrative hypothetical values and should be estimated or calibrated for a particular application rather than assumed to apply to every individual or pressure range.
Autonomic Dynamics and Cardiovascular Outputs
The autonomic response does not have to reach its target immediately. A simple representation uses first-order adjustment, with separate time constants for sympathetic and parasympathetic responses:
Here \(\tau_s\) and \(\tau_v\) are positive time constants in minutes. If pressure changes as a step and remains at its new value, the solution for either state \(z\), with target \(z_{\infty}\), time constant \(\tau\), and initial value \(z(0)\), is
For the examples, use the following hypothetical local output model, with heart rate in beats/min, stroke volume in mL/beat, and systemic vascular resistance in mmHg min/L:
These equations encode increased heart rate from sympathetic drive and reduced heart rate from parasympathetic drive. In this simplified model, sympathetic drive also increases stroke volume and systemic vascular resistance. Cardiac output is calculated from heart rate and stroke volume:
The model assumes that baseline values are \(\mathrm{HR}_0=70\) beats/min, \(\mathrm{SV}_0=70\) mL/beat, and \(\mathrm{SVR}_0=18\) mmHg min/L, with \(s=v=0\). It also assumes that the output relations are adequate over the perturbations considered. In particular, it omits direct effects of pressure on the heart, changes in venous return, and other feedback pathways. The earlier material on venous return and cardiac function treats operating-point interactions; the present output equations should not be taken as a replacement for those relationships.
Worked Example: The First Minute After a Pressure Fall
Suppose, hypothetically, that pressure is abruptly held 10 mmHg below baseline, so \(\Delta\mathrm{MAP}=-10\) mmHg, and that \(s(0)=v(0)=0\). Let \(\tau_s=4\) min and \(\tau_v=0.5\) min. Calculate the autonomic states and cardiovascular outputs at \(t=1\) min.
The sensor signal and autonomic targets are
Using the step-response equation,
The faster parasympathetic state has moved closer to its target by one minute than the sympathetic state has. Substituting these values into the output equations gives
Therefore,
The baseline cardiac output is \(70\times70/1000=4.900\) L/min. This calculation illustrates a transient response to a pressure that is held fixed; it does not yet allow the resulting cardiac output or vascular resistance to feed back and change pressure.
Steady Response and Local Sensitivities
For a sustained pressure deviation, the first-order states approach their targets. At steady state, \(s=s_{\infty}=-0.8r\) and \(v=v_{\infty}=0.5r\). Since \(r=\Delta\mathrm{MAP}/25\), the states can also be written directly in terms of pressure deviation:
The pressure sensitivities of the outputs follow by substitution. The heart-rate equation includes both autonomic contributions; omitting the vagal term would give the wrong sensitivity.
These slopes are local: for example, \(-0.544\) has units of beats/min per mmHg. A positive pressure change predicts a lower heart rate and lower resistance in this model; a negative change predicts the opposite. The sign of the stroke-volume slope is likewise negative, because increased pressure reduces sympathetic drive in the model.
Worked Example: Sustained Pressure Decrease With Pressure Clamped
Suppose a hypothetical experiment maintains \(\Delta\mathrm{MAP}=-10\) mmHg long enough for the autonomic states to reach steady state. The sensor signal remains \(r=-0.4\), so \(s=0.32\) and \(v=-0.20\). The outputs are
The corresponding cardiac output is
The predicted increases in heart rate, stroke volume, and resistance oppose the pressure decrease. The pressure clamp is an important condition: these values describe the response at a prescribed pressure, not the pressure that would result when the outputs themselves influence arterial pressure.
Closing a Simple Pressure-Feedback Loop
A local closed-loop calculation can show how the autonomic response attenuates an imposed pressure disturbance. To keep the calculation explicit, suppose that near baseline the pressure perturbation \(d\) and output changes satisfy the hypothetical linearized relation
Here \(d\) is an externally imposed pressure disturbance in mmHg; the coefficients are assumed local pressure sensitivities, with units that make each product a pressure change. This is a calibrated linearization for illustration, not an identity that must hold in every circulation. It assumes that the pressure response to the outputs can be approximated by their first-order effects and that the autonomic response is at steady state.
To obtain the cardiac-output sensitivity, differentiate \(\mathrm{CO}=\mathrm{HR}\times\mathrm{SV}/1000\) at the baseline. The first-order change is
The approximation drops the product of the two changes, \(\Delta\mathrm{HR}\Delta\mathrm{SV}/1000\), which is second order for small perturbations. Substituting the steady-state sensitivities gives
The combined output-feedback contribution to pressure is therefore
A negative combined slope means the outputs oppose a pressure deviation. Substituting the output sensitivities into the pressure equation yields a scalar equation for the closed-loop response:
The denominator is positive and greater than one, so the response has the same sign as the disturbance but a smaller magnitude. In this local static model, that is the quantitative meaning of negative feedback.
Worked Example: Closed-Loop Response to a Pressure Disturbance
Use the same hypothetical model, but now let the external disturbance be \(d=-10\) mmHg and allow the outputs to feed back. The closed-loop pressure deviation is
The pressure decrease is attenuated relative to the \(-10\) mmHg imposed disturbance. Applying the sensitivities to this closed-loop pressure change gives
Thus the predicted values are approximately \(72.80\) beats/min for heart rate, \(70.66\) mL/beat for stroke volume, \(18.099\) mmHg min/L for resistance, and \(5.142\) L/min for cardiac output when the baseline \(4.900\) L/min is increased by the first-order change \(0.242\) L/min. The calculated output changes are local linear approximations, so recomputing cardiac output from the rounded heart-rate and stroke-volume predictions can differ slightly: that multiplication includes a second-order term deliberately omitted in the linearization.
Assumptions, Diagnostics, and Alternatives
The calculations rely on several assumptions that should be checked before interpreting them. The pressure-to-sensor mapping and autonomic gains are treated as linear over the modeled range; the output equations use constant local sensitivities; baseline values are fixed; and the closed-loop example treats autonomic states as equilibrated. The transient example instead assumes a pressure step and first-order state adjustment. These are distinct conditions, so a transient response should not be calculated using steady-state autonomic values unless enough time has passed relative to the time constants.
Useful diagnostics are primarily structural and qualitative. Check that a pressure fall produces increased sympathetic drive and reduced vagal drive, and that the resulting heart-rate and resistance changes oppose the fall. Check units in each equation, especially when multiplying cardiac output by resistance to obtain a pressure change. For a linear negative-feedback model of the form \(\Delta\mathrm{MAP}=d+K\Delta\mathrm{MAP}\), the closed-loop solution is \(\Delta\mathrm{MAP}=d/(1-K)\). Here \(K=-0.9408\), so the denominator is \(1.9408\). If instead the feedback sign were positive and \(K\) approached or exceeded one, the static solution could become very large or fail to represent a stable operating point; that would signal that the assumed model or operating range needs reconsideration.
A common mistake is to compute the heart-rate change from the sympathetic term alone. In this model, \(\Delta\mathrm{HR}=12s-8v\), so both terms must be included. At steady state this gives \([12(-0.032)-8(0.020)]\Delta\mathrm{MAP}=-0.544\Delta\mathrm{MAP}\), not \(-0.384\Delta\mathrm{MAP}\). That error propagates into cardiac output and the closed-loop denominator. Another pitfall is to treat the product \(\mathrm{HR}\times\mathrm{SV}\) as exactly linear in changes; its first-order approximation is appropriate for small deviations, but larger changes require retaining the product term or using a nonlinear model.
A static sigmoid model of baroreceptor firing can represent saturation and a finite operating range, but by itself it does not describe response timing. The first-order dynamic model used here adds timing with minimal complexity, while still omitting delays, multiple neural pathways, and adaptation. A more detailed model could include nonlinear sensor-response curves, distinct cardiac and vascular sympathetic states, explicit delays, and additional cardiovascular compartments. Those additions are warranted when data show saturation, marked time dependence, or outputs that cannot be represented by a single shared sympathetic state. No reduced model should be extrapolated far beyond the pressure range over which its gains and output relations are defensible.
The baroreflex is only one contributor to vascular and organ-level control. The present model treats systemic vascular resistance as a lumped output; it does not resolve local control of individual organ blood flow.
Check Your Understanding
Answer each question, then compare with the solutions below.
- For a sustained pressure increase of 5 mmHg, calculate \(r\), \(s\), and \(v\) at steady state.
- For that same pressure increase, calculate the model-predicted changes in heart rate and systemic vascular resistance.
- With \(\tau_s=4\) min, a target \(s_{\infty}=0.16\), and \(s(0)=0\), calculate \(s(4)\).
- If a linear closed-loop model has \(d=6\) mmHg and feedback slope \(K=-0.5\) in \(\Delta\mathrm{MAP}=d+K\Delta\mathrm{MAP}\), calculate \(\Delta\mathrm{MAP}\). Does the feedback attenuate or amplify the disturbance?
Solutions. (1) \(r=5/25=0.20\), \(s=-0.8(0.20)=-0.16\), and \(v=0.5(0.20)=0.10\). (2) \(\Delta\mathrm{HR}=12(-0.16)-8(0.10)=-2.72\) beats/min, and \(\Delta\mathrm{SVR}=0.6(-0.16)=-0.0960\) mmHg min/L. (3) \(s(4)=0.16(1-\exp(-4/4))=0.16(1-0.3679)\approx0.1011\). (4) \(\Delta\mathrm{MAP}=6/(1-(-0.5))=6/1.5=4.00\) mmHg. The feedback attenuates the positive disturbance because the resulting pressure deviation is smaller than 6 mmHg.